MATH1014 Tutorial 4
Integration by Parts
ˆ
ˆ
u(x)v 0 (x) dx = u(x)v(x) −
• For indefinite integrals,
ˆ
ˆ
b
0
• For definite integrals,
ˆ
v(x)u0 (x) dx, or the compact form
u(x)v (x) dx =
[u(x)v(x)]ba
v du.
b
v(x)u0 (x) dx.
−
a
ˆ
u dv = uv −
a
• Usually we should choose u(x) to be a function that becomes simpler when differentiated, and v 0 (x) to be a
function with antiderivative immediately available.
ˆ
x tan2 x dx.
Example 1. Evaluate
First we rewrite tan2 x as sec2 x − 1, since sec2 x has a readily available antiderivative. Thus
ˆ
ˆ
x tan2 x dx = x(sec2 x − 1) dx
ˆ
ˆ
2
= x sec x dx − x dx
ˆ
x2
= x d tan x −
2
ˆ
2
x
= x tan x −
− tan x dx
2
x2
= x tan x −
+ ln | cos x| + C.
2
ˆ
√
3
tan−1
Example 2. Evaluate
1
1
dx.
x
Applying integration by parts directly, we get
ˆ
√
3
tan
−1
1
√3 ˆ √3
1
1
−1 1
−
x d tan−1
dx = x tan
x
x 1
x
1
ˆ √3
√
x
1
=
3 tan−1 √ − tan−1 1 +
dx
2
x +1
3
1
√
ˆ √3
3π π 1
d(x2 + 1)
=
− +
6
4
2 1
x2 + 1
√
√3
3π π 1 =
− +
ln |x2 + 1| 1
4
2
√6
3π π ln 2
=
− +
.
6
4
2
Prepared by Leung Ho Ming
Homepage: http://ihome.ust.hk/~malhm
1
Example 3. For a 6= 0, derive the reduction formulas
ˆ
ˆ
ˆ
ˆ
xn sin ax n
xn cos ax n
n
n−1
n
x cos ax dx =
x
xn−1 cos ax dx.
−
sin ax dx and
x sin ax dx = −
+
a
a
a
a
ˆ
Hence evaluate
x3 sin 2x dx.
Using integration by parts,
ˆ
ˆ
1
n
xn d sin ax
x cos ax dx =
a
ˆ
xn sin ax 1
=
sin ax dxn
−
a
a
ˆ
xn sin ax n
xn−1 sin ax dx,
=
−
a
a
ˆ
ˆ
1
xn d cos ax
x sin ax dx = −
a
ˆ
xn cos ax 1
=−
cos ax dxn
+
a
a
ˆ
xn cos ax n
xn−1 cos ax dx.
=−
+
a
a
n
and
Hence
ˆ
x3 cos 2x 3
+
x2 cos 2x dx
2
2
ˆ
x3 cos 2x 3 x2 sin 2x 2
x sin 2x dx
=−
+
−
2
2
2
2
ˆ
x3 cos 2x 3 x2 sin 2x
x cos 2x 1
=−
+
− −
+
cos 2x dx
2
2
2
2
2
x3 cos 2x 3x2 sin 2x 3x cos 2x 3
+
+
− sin 2x + C.
=−
2
4
4
8
ˆ 1
Example 4. Let In denote the integral
(1 − x2 )n dx, where n ≥ 1 is a positive integer. Show that In =
ˆ
x3 sin 2x dx = −
0
22n (n!)2
, where n! = 1 · 2 · 3 · · · (n − 1) · n is the factorial of n.
then deduce that In =
(2n + 1)!
Using integration by parts,
ˆ 1
ˆ 1
In =
(1 − x2 )n dx = [x(1 − x2 )n ]10 −
x d(1 − x2 )n
0
ˆ
0
1
(−x2 )(1 − x2 )n−1 dx
= −2n
0
ˆ
1
(1 − x2 − 1)(1 − x2 )n−1 dx
= −2n
0
= −2n(In − In−1 )
= 2nIn−1 − 2nIn .
2n
In−1 . From the starting case I1 =
Hence (2n + 1)In = 2nIn , and In =
2n + 1
In =
ˆ
1
(1 − x2 ) dx =
0
2n
2n
2n − 2
In−1 =
·
In−2
2n + 1
2n + 1 2n − 1
2n
2n − 2
4
=
·
· · · · · I1
2n + 1 2n − 1
5
2n
2n − 2
4 2
=
·
· ··· · ·
2n + 1 2n − 1
5 3
(2n)2 (2n − 2)2 · · · (4)2 (2)2
=
(2n + 1)(2n) · · · (5)(4)(3)(2)
22n (n!)2
=
.
(2n + 1)!
2
2
,
3
2n
In−1 ,
2n + 1
Trigonometric Integrals
ˆ
• For integrals in the form
sinm x cosn x dx:
– If m is odd, split off sin x and use u = cos x.
– If n is odd, split off cos x and use u = sin x.
– If m and n are both even, use the identities sin2 x =
ˆ
• For integrals in the form tanm x secn x dx:
1−cos 2x
, cos2
2
x=
1+cos 2x
, sin x cos x
2
=
sin 2x
2 .
– If m is odd, split off sec x tan x and use u = sec x.
– If n is even, split off sec2 x and use u = tan x.
– If
´ m nis even and n is odd, rewrite the even powers of tan x in terms of sec x and use reduction formula of
sec x dx.
ˆ
ˆ
ˆ
• For integrals in the form sin mx cos nx dx, sin mx sin nx dx or cos mx cos nx dx:
– Apply the product-to-sum identities
sin A cos B =
sin(A−B)+sin(A+B)
, sin A sin B
2
ˆ
=
cos(A−B)−cos(A+B)
, cos A cos B
2
=
cos(A−B)+cos(A+B)
.
2
π/2
sin5 x dx.
Example 5. Evaluate
0
Let u = cos x, then
ˆ
ˆ
π/2
0
sin5 x dx = −
0
ˆ
(1 − u2 )2 du
1
1
(1 − 2u2 + u4 ) du
=
0
1
1
2
= u − u3 + u5
3
5
0
8
=
.
15
ˆ
π
sin2 t cos4 t dt.
Example 6. Evaluate
0
Applying identities, we get
ˆ
ˆ
π
2
π
4
sin t cos t dt =
0
0
ˆ
=
=
=
=
=
1 − cos 2t
·
2
1 + cos 2t
2
2
dt
1 π
(1 − cos2 2t)(1 + cos 2t) dt
8 0
ˆ
1 π
(sin2 2t + sin2 2t cos 2t) dt
8 0
ˆ π
ˆ
1
1 − cos 4t
1 π 2
dt +
sin 2t d sin 2t
8
2
2 0
0
π 3 π 1
sin 4t
sin 2t
t−
+
16
4
3
0
0
π
.
16
3
Example 7. Suppose m and n are positive integers, show that
(
ˆ
π
sin mx sin nx dx =
−π
If m 6= n,
ˆ π
sin mx sin nx dx =
−π
1
2
ˆ
0, m 6= n
.
π, m = n
π
[cos(m − n)x − cos(m + n)x] dx =
−π
π
1 sin(m − n)x sin(m + n)x
−
= 0,
2
m−n
m+n
−π
since m + n and m − n are nonzero integers, and sin kπ = 0 for all integers k.
If m = n, the integral becomes
π
ˆ
ˆ π
1
sin 2nx
1 π
2
(1 − cos 2nx) dx =
sin nx dx =
x−
= π.
2 −π
2
2n
−π
−π
4