Trigonometric Substitution
Lecture 9
√
√
√
Section 8.4 Integrals Involving a2 − x 2 , a2 + x 2 , x 2 − a2
Trigonometric Substitutions
Jiwen He
Department of Mathematics, University of Houston
[email protected]
http://math.uh.edu/∼jiwenhe/Math1432
Z p
a2 − x 2 dx,
Jiwen He, University of Houston
Z p
a2 + x 2 dx,
Z p
x 2 − a2 dx
Math 1432 – Section 26626, Lecture 9
February 12, 2008
1 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
1−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
√
1 − sin2 u
x
dx
1 − x2
1 − x2
= cos2 u
= sin u
= cos u du
= cos2 u
Example
Z p
Z √
Z
1 − x 2 dx =
cos2 u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1
=
sin−1 x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
2 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2
=
sin−1 x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
√
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z p
Z √
Z
2
2
2
2
2
a − x dx =
a cos u a cos u du = a
cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
=
(u + sin u cos u) + C
2
p
a2 −1
=
sin x + x 1 − x 2 + C
2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
3 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Area of a Circle
√
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z r p
Z
2
2
r − x dx =
π
2
√
r 2 cos2 u r
=r
2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
cos u du = r
− π2
−r
a2 − x 2
1
1
u + sin 2u
2
4
Z
π
2
cos2 u du
− π2
π
2
=
− π2
2
πr 2
2
= area enclosed by a semicircle of radius r
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
4 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u
x −c
dx
2
a − (x − c)2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Example
Z q
Z p
2 − x 2 + 4x dx =
2 + 4 − (x 2 + 4x + 4) dx
Z q
Z
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
q
x −2
−1 x − 2
√ +
= 3 sin
6 − (x − 2)2 + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
5 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
√
Trigonometric Substitution
Sine Substitution:
a2
−
Sine Substitution Tangent Substitution Secant Substitution
x2
Sine Substitution:
1 − sin2 u
x
dx
a2 − x 2
Example
Z √
a2 − x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
=2
du = 2 csc u − sin u du
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
= 2 ln | −
| + 4 − x2 + C = · · ·
x
x
4 − x2
dx =
x
Jiwen He, University of Houston
√
Math 1432 – Section 26626, Lecture 9
February 12, 2008
6 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
Trigonometric Substitution
Tangent Substitution: 1 + x
Sine Substitution Tangent Substitution Secant Substitution
2
Tangent Substitution: 1 + x 2
1 + tan2 u
x
dx
1 + x2
= sec2 u
= tan u
= sec2 u du
= sec2 u
Example
Z
Z
Z
1
1
2
dx
=
sec
u
du
=
cos2 u du
4u
(1 + x 2 )2
sec
Z 1
1 1
1
+ cos 2u du = u + sin 2u + C
=
2 2
2
4
1
= (u + sin u cos u) + C
2
1
x
1
−1
=
tan x + √
·√
+ C = ···
2
1 + x2
1 + x2
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
7 / 16
√
Trigonometric Substitution
Tangent Substitution:
a2
Sine Substitution Tangent Substitution Secant Substitution
+ x2
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z √
Z
a2 + x 2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x 2 x a2
a2 + x 2 x
=
+ ln + + C = ···
2
a
a
a
a
=
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
8 / 16
√
Trigonometric Substitution
Tangent Substitution:
a2
Sine Substitution Tangent Substitution Secant Substitution
+ x2
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z √
Z
a2 + x 2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x 2 x a2
a2 + x 2 x
=
+ ln + + C = ···
2
a
a
a
a
=
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
8 / 16
√
Trigonometric Substitution
Tangent Substitution:
a2
Sine Substitution Tangent Substitution Secant Substitution
+ x2
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z √
Z
a2 + x 2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x 2 x a2
a2 + x 2 x
=
+ ln + + C = ···
2
a
a
a
a
=
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
8 / 16
√
Trigonometric Substitution
Tangent Substitution:
a2
Sine Substitution Tangent Substitution Secant Substitution
+ x2
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z √
Z
a2 + x 2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x 2 x a2
a2 + x 2 x
=
+ ln + + C = ···
2
a
a
a
a
=
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
8 / 16
√
Trigonometric Substitution
Tangent Substitution:
a2
Sine Substitution Tangent Substitution Secant Substitution
+ x2
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z √
Z
a2 + x 2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x 2 x a2
a2 + x 2 x
=
+ ln + + C = ···
2
a
a
a
a
=
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
8 / 16
√
Trigonometric Substitution
Tangent Substitution:
a2
Sine Substitution Tangent Substitution Secant Substitution
+ x2
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z √
Z
a2 + x 2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x 2 x a2
a2 + x 2 x
=
+ ln + + C = ···
2
a
a
a
a
=
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
8 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
a2 + x 2
Example
Z
1
1
dx =
2
x 9 + 4x 2
√
Jiwen He, University of Houston
Z
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
1
q
3
sec2 u du
2
3
9
tan u 4 sec2 u
Z 2
Z
1
sec u
1
du =
=
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x 2 − 3
= ln |
|+C
3
2x
Math 1432 – Section 26626, Lecture 9
February 12, 2008
9 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example
Tangent Substitution:
1 + tan2 u
x
dx
2
a + x2
√
a2 + x 2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z
Z
Z
x2
4 tan2 u
8 tan2 u sec2 u
2
√
√
√
du
dx =
2 sec u du =
4 + x2
4 + 4 tan2 u
4 sec2 u
Z
Z
8 tan2 u sec2 u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
10 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Tangent Substitution:
1 + tan2 u
x −c
dx
2
a + (x − c)2
p
a2 + (x − c)2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z q
2
x 16 + x − 6x dx = x 16 − 9 + (x 2 − 6x + 9) dx
Z q
= x 7 + (x − 3)2 dx
Z √
√
√
=
7 tan u + 3
7 sec2 u 7 sec2 u du
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
11 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Tangent Substitution:
1 + tan2 u
x −c
dx
2
a + (x − c)2
p
a2 + (x − c)2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z q
2
x 16 + x − 6x dx = x 16 − 9 + (x 2 − 6x + 9) dx
Z q
= x 7 + (x − 3)2 dx
Z √
√
√
=
7 tan u + 3
7 sec2 u 7 sec2 u du
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
11 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Tangent Substitution:
1 + tan2 u
x −c
dx
2
a + (x − c)2
p
a2 + (x − c)2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z q
2
x 16 + x − 6x dx = x 16 − 9 + (x 2 − 6x + 9) dx
Z q
= x 7 + (x − 3)2 dx
Z √
√
√
=
7 tan u + 3
7 sec2 u 7 sec2 u du
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
11 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Tangent Substitution:
1 + tan2 u
x −c
dx
2
a + (x − c)2
p
a2 + (x − c)2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z q
2
x 16 + x − 6x dx = x 16 − 9 + (x 2 − 6x + 9) dx
Z q
= x 7 + (x − 3)2 dx
Z √
√
√
=
7 tan u + 3
7 sec2 u 7 sec2 u du
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
11 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Tangent Substitution:
1 + tan2 u
x −c
dx
2
a + (x − c)2
p
a2 + (x − c)2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z q
2
x 16 + x − 6x dx = x 16 − 9 + (x 2 − 6x + 9) dx
Z q
= x 7 + (x − 3)2 dx
Z √
√
√
=
7 tan u + 3
7 sec2 u 7 sec2 u du
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
11 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Tangent Substitution:
1 + tan2 u
x −c
dx
2
a + (x − c)2
p
a2 + (x − c)2
= sec2 u
= a tan u
= a sec2 u du
= a2 sec2 u
Example
Z p
Z q
2
x 16 + x − 6x dx = x 16 − 9 + (x 2 − 6x + 9) dx
Z q
= x 7 + (x − 3)2 dx
Z √
√
√
=
7 tan u + 3
7 sec2 u 7 sec2 u du
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
11 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
2
Secant Substitution: x − 1
Secant Substitution: x 2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
2
x − 1 = tan2 u
Example
Z
Z
Z
1
sec u tan u
1
√
√
dx =
du =
du
2
2
2
2
sec u
x x −1
Z sec u tan u
= cos u du = sin u + C
√
x2 − 1
=
+C
x
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
12 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
√
Trigonometric Substitution
Secant Substitution:
x2
−
Sine Substitution Tangent Substitution Secant Substitution
a2
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x 2 − a2 = a2 tan2 u
Example
Z
Z
2 sec u tan u
1
1
=
du
du
2u
2 u)3/2
8
tan
2
sec
u(4
tan
Z
Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
x(x 2 − 4)3/2
Jiwen He, University of Houston
Z
Math 1432 – Section 26626, Lecture 9
February 12, 2008
13 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example
Z
Z
x
x
√
p
dx
dx =
2
2
x
+
2x
−
3
(x
+
2x
+
1)
−
316
−
1
Z
x
p
=
, dx
(x + 1)2 − 4
Z
2 sec u − 1
√
=
2 sec u tan u du
4 tan2 u
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
14 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example
Z
Z
x
x
√
p
dx
dx =
2
2
x
+
2x
−
3
(x
+
2x
+
1)
−
316
−
1
Z
x
p
=
, dx
(x + 1)2 − 4
Z
2 sec u − 1
√
=
2 sec u tan u du
4 tan2 u
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
14 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example
Z
Z
x
x
√
p
dx
dx =
2
2
x
+
2x
−
3
(x
+
2x
+
1)
−
316
−
1
Z
x
p
=
, dx
(x + 1)2 − 4
Z
2 sec u − 1
√
=
2 sec u tan u du
4 tan2 u
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
14 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example
Z
Z
x
x
√
p
dx
dx =
2
2
x
+
2x
−
3
(x
+
2x
+
1)
−
316
−
1
Z
x
p
=
, dx
(x + 1)2 − 4
Z
2 sec u − 1
√
=
2 sec u tan u du
4 tan2 u
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
14 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example
Z
Z
x
x
√
p
dx
dx =
2
2
x
+
2x
−
3
(x
+
2x
+
1)
−
316
−
1
Z
x
p
=
, dx
(x + 1)2 − 4
Z
2 sec u − 1
√
=
2 sec u tan u du
4 tan2 u
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
14 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Example: Completing the Square
Secant Substitution:
√
x 2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example
Z
Z
x
x
√
p
dx
dx =
2
2
x
+
2x
−
3
(x
+
2x
+
1)
−
316
−
1
Z
x
p
=
, dx
(x + 1)2 − 4
Z
2 sec u − 1
√
=
2 sec u tan u du
4 tan2 u
Finish it!
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
14 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Summary
a2 − x 2
1 − sin2 u
x
dx
a2 − x 2
a2 + x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
1 + tan2 u
x
dx
2
a + x2
x 2 − a2
= sec2 u
sec2 u − 1 = tan2 u
= a tan u
x = a sec u
= a sec2 u du
dx = a sec u tan u d
= a2 sec2 u
x 2 − a2 = a2 tan2 u
√
x
sin u =
a
√
a2 − x 2
a
x
tan u = √
a2 − x 2
cos u =
Jiwen He, University of Houston
x
sin u = √
2
a + x2
a
cos u = √
2
a + x2
x
tan u =
a
Math 1432 – Section 26626, Lecture 9
sin u =
cos u =
tan u =
a
x√
x 2 − a2
x
x 2 − a2
a
February 12, 2008
15 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Summary
a2 − x 2
1 − sin2 u
x
dx
a2 − x 2
a2 + x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
1 + tan2 u
x
dx
2
a + x2
x 2 − a2
= sec2 u
sec2 u − 1 = tan2 u
= a tan u
x = a sec u
= a sec2 u du
dx = a sec u tan u d
= a2 sec2 u
x 2 − a2 = a2 tan2 u
√
x
sin u =
a
√
a2 − x 2
a
x
tan u = √
a2 − x 2
cos u =
Jiwen He, University of Houston
x
sin u = √
2
a + x2
a
cos u = √
2
a + x2
x
tan u =
a
Math 1432 – Section 26626, Lecture 9
sin u =
cos u =
tan u =
a
x√
x 2 − a2
x
x 2 − a2
a
February 12, 2008
15 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Summary
a2 − x 2
1 − sin2 u
x
dx
a2 − x 2
a2 + x 2
= cos2 u
= a sin u
= a cos u du
= a2 cos2 u
1 + tan2 u
x
dx
2
a + x2
x 2 − a2
= sec2 u
sec2 u − 1 = tan2 u
= a tan u
x = a sec u
= a sec2 u du
dx = a sec u tan u d
= a2 sec2 u
x 2 − a2 = a2 tan2 u
√
x
sin u =
a
√
a2 − x 2
a
x
tan u = √
a2 − x 2
cos u =
Jiwen He, University of Houston
x
sin u = √
2
a + x2
a
cos u = √
2
a + x2
x
tan u =
a
Math 1432 – Section 26626, Lecture 9
sin u =
cos u =
tan u =
a
x√
x 2 − a2
x
x 2 − a2
a
February 12, 2008
15 / 16
Trigonometric Substitution
Sine Substitution Tangent Substitution Secant Substitution
Outline
Trigonometric Substitution
Sine Substitution
Tangent Substitution
Secant Substitution
Jiwen He, University of Houston
Math 1432 – Section 26626, Lecture 9
February 12, 2008
16 / 16