Today: 6.2 Trig substitution
Warm up:
1. Calculate the following integrals.
Z
Z
2
(a)
cos (x) dx (b)
cos2 (x) sin2 (x) dx
Z
Z
2
(c)
cot (x) dx (d)
tan3 (x) dx
2. Simplify the following expressions.
(a) sin(cos
1
(x))
(b) tan(sec
1
(c) sin(2 cos
1
(x))
(d) cos(2 cos
(x))
1
(x))
(1a)
Z
2
cos (x) dx =
1
2
Z
1 + cos(2x) dx
= 12 (x + 12 sin(2x)) + C.
Z
Z
2
2
(1b)
cos (x) sin (x) dx = ( 12 sin(2x))2 dx
Z
Z
2
1
1
= 4 sin (2x) dx = 8 1 cos(4x) dx
= 18 (x
1
4
sin(4x)) + C =
Z
Z
2
1
(1c)
cot (x) dx = 2 csc2 (x)
(1d)
Z
1
32 (4x
sin(4x)) + C.
1 dx
=
cot(x) x + C
Z
tan3 (x) dx = tan(x)(sec2 (x) 1) dx
Z
Z
2
=
tan(x) sec (x) dx
tan(x) dx
|
{z
}
| {z }
=u du, with u=tan(x)
1
2
tan2 (x) + ln | cos(x)| + C.
u
1
du, with u=cos(x)
(2a) sin(cos 1 (x)): Let u = cos
use the triangle
1
u
x
So sin(cos
1 (x))
1 (x))
p
= tan(u) =
x2
x2 .
1
1 (x)
p
so that cos(u) = x. Thus
1
p
= sin(u) =
(2b) tan(sec 1 (x)): Let u = sec
use the triangle
x
u
1
So tan(sec
1 (x)
so that sec(u) = x. Thus
x2
p
1
1
x2 .
(2c) sin(2 cos 1 (x)): Let u = cos 1 (x) so that cos(u) = x. Thus
again use the triangle
p
1
1 x2
u
x
Thus
p
sin(2 cos 1 (x)) = sin(2u) = 2 sin(u) cos(u) = 2x 1 x2 .
(2d) cos(2 cos 1 (x)): Using the same substitution and triangle as
above, we have
cos(2 cos
1
(x)) = cos(2u) = (cos(u))2
= x2
(1
x2 ) = 2x2
1.
(sin(u))2
Trig substitution, or “reverse u-sub”
Normal straightforward u-sub:
Calculate
Z p
x 1
x2 dx.
Let u = 1 x2 . Then du = 2x dx. So
Z p
Z
1
x 1 x2 dx = 2 u1/2 dx = 12 23 u3/2 +C =
1
3 (1
x2 )3/2 +C.
A little more sophistication: Calculate
Z p
e x dx.
p
1
Let u = x, so that du = 2p1 x dx = 2u
dx, and thus dx = 2u du.
So
Z p
Z
Z
x
u
u
e dx = e (2u) du = 2ue
2 eu du
(integration by parts with f (u) = 2u and g 0 (u) = eu )
p
p p
= 2 xe x 2e x + C.
Trig substitution, or “reverse u-sub”
How about
Z p
1
x2 dx?
p
We could try lettingpu = 1 x2 , so that x = 1 u. Further,
du = 2x dx = 2 1 u du. So
p
Z p
Z
Z r
u
u
p
1 x2 dx = 12
du = 12
du . . .
1 u
1 u
Trig substitution, or “reverse u-sub”
How about
Z p
1
x2 dx?
Instead of using a substitution that looks like u = f (x), we can try
making a substitution that looks like x = f (u). Is there a function
f (u) such that 1 f 2 (u) is a perfect square? Think
cos2 (u) + sin2 (u) = 1.
Let x = cos(u). Then dx = sin(u) du. So
Z p
Z p
1 x2 dx =
1 cos2 (u) · ( sin(u)) du
Z q
Z
=
sin2 (u) · sin(u) du =
sin2 (u) du
Z
1
= 2 cos(2u) 1 du = 12 ( 12 sin(2u) u) + C
=
1
4
sin(2 cos
1
1
2
(x)))
cos
1
(x) + C.
Simplifying sin(2 sin 1 (x)))
So we don’t get too bogged down, let’s go back to writing this as
sin(2u), where cos(u) = x so that u = cos
1
(x).
First,
sin(2u) = 2 sin(u) cos(u).
Next, use the triangle
p
1
u
x
So
Z p
1
1
x2
p
sin(2u) = 2 sin(u) cos(u) = 2 1
x2 dx = · · · =
= 12 x
p
1
1
2
cos
x2
x2 · x.
1
(x) + 14 sin(2 sin
1
2
cos
1
(x) + C
1
(x))) + C
Check against geometry:
Rp
Note: y =
p
1
2
⇣ p
x 1
⌘
1
1
x2 is the upper half of the graph of y 2 + x2 = 1:
dx =
x2
1 (x)
x2
1
cos
+C
h
u
a
Recall: the area of a wedge with angle u of a circle or radius r is
A = (u/2⇡)⇡r2 = 12 ur2 .
R1p
So, for example, the integral I = a 1 x2 dx should be
(area of the wedge) (area of the triangle) = 12 u
p
Since h = 1 a2 and u = cos 1 (a), we have
p
I = 12 cos 1 (a) 12 a 1 a2 .
1
2 ah.
Check against geometry:
Rp
1
x2
dx =
1
2
⇣ p
x 1
x2
1
cos
1 (x)
⌘
+C
h
u
a
Geometrically,
Z 1p
1 x2 dx = 12 u
a
1
2 ah
=
1
2
cos
1
(a)
1
2a
p
1
Checking against the formula we computed:
Z 1p
⇣ p
⌘
1
1 x2 dx = 12 x 1 x2 cos 1 (x) x=a
a
⇣
⌘
p
1
1
1
2
= 2 (1 · 0 cos (1)) (a 1 a
cos (a))
p
= 12 cos 1 (a) 12 a 1 a2 X (cos 1 (1) = 0).
a2 .
You try:
Calculate the following integrals using the suggested substitution.
Be sure to simplify your answers.
R p 1 x2
1.
dx using x = sin(u)
x2
2.
3.
R
R
p1
x2 1+x2
p x
1+x2
dx using x = tan(u)
dx two ways:
(a) Let x = tan(u)
(b) Let u = 1 + x2 .
You try:
Calculate the following integrals using the suggested substitution.
Be sure to simplify your answers.
R p 1 x2
1.
dx using x = sin(u): dx = cos(u) du, so
x2 p
R 1 sin2 (u)
R cos2 (x)
I=
·
cos(u)
du
=
du
2
sin (u)
sin2 (u)
R
= cot2 (u) du = cot(u) u + C p
1 x2
= cot(sin 1 (x)) sin 1 (x) + C=
sin 1 (x) + C
x
R
2. x2 p11+x2 dx using x = tan(u): dx = sec2 (u) du, so
R
R
sec2 (u)
1
2 (u) du =
p
I=
sec
du
2
tan (u) sec(u)
tan2 (u) 1+tan2 (u)
R cos(u)
= sin
sin 1 (u) + C = csc(tan 1 (x)) + C
2 (u) du =
3.
R
=
p x
1+x2
dx two ways:
(a) Let x = tan(u)
=
p
p
1+x2
x
+ C.
(b) Let u = 1 + x2 .
1 + x2 + C
You try:
Calculate the following integrals using the suggested substitution.
Be sure to simplify your answers.
R x
3. p1+x
dx two ways:
2
(a) Let x = tan(u): dx = sec2 (u) du, so that
Z
Z
x
tan(u)
p
p
dx =
sec2 (u) du
2
2
1+x
1 + tan (x)
Z
p
= sec(u) tan(u) du = sec(u) + C = 1 + x2 + C
p
1 + x2
u
1
x
(b) Let u = 1 + x2 : Then du = 2x dx, so that
Z
Z
p
x
1
p
dx = 2 u 1/2 du = 12 ·2u1/2 +C = 1 + x2 +C.
1 + x2
Another geometric example
Compute the area of an ellipse with minor radius 3 and major
radius 4.
y2
x2
+
=1
2
4
32
The desired
area is half the area under
p
R 4 pthe curve
y = 3 1 (x/4)2 . So A = 2 · 3 4 1 (x/4)2 dx.
Let x/4 = sin(u), so that dx = 4 cos(u) du. Thus,
Z p
Z q
Z
2
1 x2 /42 dx =
1 sin (u) 4 cos(u) du = 4 cos2 (u) du
=
4· 12 (u+cos(u) sin(u))+C
⇣
= 2 sin
1
p
(x/4) + (x/4) 1
(x/4)2
Another geometric example
Compute the area of an ellipse with minor radius 3 and major
radius 4.
2
x2
+ y32 = 1
42
The desired
area is half the area under the curve
p
y = 3 1 (x/4)2 . So Z 4
p
A=6
1 (x/4)2 dx
4
⇣ ⇣
⌘⌘
p
4
1
2
= 6 2 sin (x/4) + (x/4) 1 (x/4)
x= 4
p
p
= 12 sin 1 (1) + 1 1
sin 1 (x 1) + ( 1) 1
= 12(⇡/2
( ⇡/2)) = 12⇡ .
1
⌘
+C
Completing the square
Compute
Z
Rewrite
x2
4x = x2
4x + 4
p
1
x2
4x
dx
2)2
4 = (x
4 = 4(
x 2 2
2
1).
Recall, this is called completing the square. In general, if you’re
interested in rewriting
ax2 + bx + c,
note that (x + b/2a)2 = x2 + (b/a)x + (b/2a)2 .
So
ax2 + bx + c = a x2 + (b/a)x + c = a (x + b/2a)2
=
p
a(x + b/2a)2
2
1
a (b/2)
(b/2a)2 + c
+ c.
Completing the square
Compute
Z
Rewrite
x2
4x = x2
p
1
x2
4x
4x + 4
Then
Z
Z
1
1
p
p
dx =
x2 4x
(x 2)2
dx
4 = (x
4
dx =
Z
2)2
4.
p
2 ((x
1
2)/2)2
2)/2 = sec(u). Then 12 dx = sec(u) tan(u) du, so that
Z
1
p
I=
2 sec(u) tan(u) du
2 (u)
2
sec
1
Z
sec(u) tan(u)
=
du = ln | sec(u) + tan(u)| + C
tan(u)
x 2
q
q
2
x 2 2
1
x
2
2
u
= ln 2 + ( x 2 2 )2 1 + C
Let (x
1
1
dx.
You try:
Calculate the following integrals. Be sure to simplify your answers.
Remember, the Pythagorean identities are
cos2 (u) + sin2 (u) = 1
1.
2.
3.
R
p 1
x2 1
Rp
R
3
e2x
dx
x2 dx
x
pe
1+e2x
dx
and
1 + tan2 (u) = sec2 (u).