today: 6.2 trig substitution · today: 6.2 trig substitution warm up: 1.calculate the following...

Today: 6.2 Trig substitutionWarm up:

1. Calculate the following integrals.

(a)

∫cos2(x) dx (b)

∫cos2(x) sin2(x) dx

(c)

∫cot2(x) dx (d)

∫tan3(x) dx

2. Simplify the following expressions.

(a) sin(cos−1(x)) (b) tan(sec−1(x))

(c) sin(2 cos−1(x)) (d) cos(2 cos−1(x))

Answers:

(1a) 12(x+ cos(x) sin(x)) + C (1b) 1

32(4x− sin(4x)) + C

(1c) − cot(x)− x+ C (1d) 12 tan

2(x) + ln | cos(x)|+ C

(2a)√

1− x2 (2b)√x2 − 1 (2c) 2x

√1− x2 (2d) 2x2 − 1

(1a)

∫cos2(x) dx

= 12

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


=

∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=

∫tan(x)(sec2(x)− 1) dx

=

∫tan(x) sec2(x) dx︸︷︷︸=u du, with u=tan(x)

−∫

tan(x) dx︸︷︷︸−u−1 du, with u=cos(x)

12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


=

∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)

∫cos2(x) sin2(x) dx =

∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx

= 18

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C

= 132(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx

= 12

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx = 1

2

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx

=


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(1a)

∫cos2(x) dx = 1

2

∫1 + cos(2x) dx

= 12(x+ 1

2 sin(2x)) + C.

(1b)


∫(12 sin(2x))

2 dx

= 14

∫sin2(2x) dx = 1

8

∫1− cos(4x) dx

= 18(x−

14 sin(4x)) + C = 1

32(4x− sin(4x)) + C.

(1c)

∫cot2(x) dx = 1

2

∫csc2(x)− 1 dx

= − cot(x)− x+ C

(1d)

∫tan3(x) dx =


=


−∫


12 tan

2(x) + ln | cos(x)|+ C.

(2a) sin(cos−1(x)):

Let u = cos−1(x) so that cos(u) = x. Thususe the triangle

√1− x2

x

1

u

So sin(cos−1(x)) = sin(u) =√1− x2 .

(2b) tan(sec−1(x)):

Let u = sec−1(x) so that sec(u) = x. Thususe the triangle

√x2 − 1

1

x

u

So tan(sec−1(x)) = tan(u) =√1− x2 .

(2a) sin(cos−1(x)): Let u = cos−1(x) so that cos(u) = x.

Thususe the triangle

√1− x2

x

1

u




√x2 − 1

1

x

u


(2a) sin(cos−1(x)): Let u = cos−1(x) so that cos(u) = x. Thususe the triangle

√1− x2

x

1

u




√x2 − 1

1

x

u



√1− x2

x

1

u


(2b) tan(sec−1(x)): Let u = sec−1(x) so that sec(u) = x.

Thususe the triangle

√x2 − 1

1

x

u



√1− x2

x

1

u


(2b) tan(sec−1(x)): Let u = sec−1(x) so that sec(u) = x. Thususe the triangle

√x2 − 1

1

x

u


(2c) sin(2 cos−1(x)):

Let u = cos−1(x) so that cos(u) = x. Thusagain use the triangle

√1− x2

x

1

u

Thus

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 asabove, we have

cos(2 cos−1(x)) = cos(2u)

= (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .

(2c) sin(2 cos−1(x)): Let u = cos−1(x) so that cos(u) = x.

Thusagain use the triangle

√1− x2

x

1

u

Thus


= 2 sin(u) cos(u) = 2x√1− x2 .

(2d) cos(2 cos−1(x)):



= (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .

(2c) sin(2 cos−1(x)): Let u = cos−1(x) so that cos(u) = x. Thusagain use the triangle

√1− x2

x

1

u

Thus


= 2 sin(u) cos(u) = 2x√1− x2 .

(2d) cos(2 cos−1(x)):



= (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .


√1− x2

x

1

u

Thus

sin(2 cos−1(x)) = sin(2u) = 2 sin(u) cos(u)

= 2x√1− x2 .

(2d) cos(2 cos−1(x)):



= (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .


√1− x2

x

1

u

Thus

sin(2 cos−1(x)) = sin(2u) = 2 sin(u) cos(u) = 2x√1− x2 .

(2d) cos(2 cos−1(x)):



= (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .


√1− x2

x

1

u

Thus


(2d) cos(2 cos−1(x)): Using the same substitution and triangle asabove, we have


= (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .


√1− x2

x

1

u

Thus



cos(2 cos−1(x)) = cos(2u) = (cos(u))2 − (sin(u))2

= x2 − (1− x2) = 2x2 − 1 .


√1− x2

x

1

u

Thus




= x2 − (1− x2)

= 2x2 − 1 .


√1− x2

x

1

u

Thus




= x2 − (1− x2) = 2x2 − 1 .

Trig substitution, or “reverse u-sub”Normal straightforward u-sub:Calculate ∫

x√

1− x2 dx.

Let u = 1− x2. Then du = −2x dx. So∫x√

1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.

A little more sophistication: Calculate∫e√x dx.

Let u =√x, so that du = 1

2√xdx = 1

2udx, and thus dx = 2u du.

So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du

(integration by parts with f(u) = 2u and g′(u) = eu)

= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.

Let u = 1− x2. Then du = −2x dx.

So∫x√

1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx

= −1223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C

= −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx

= 12udx, and thus dx = 2u du.

So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du

= 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.


x√

1− x2 dx.


1− x2 dx = −12

∫u1/2 dx = −1

223u

3/2+C = −13(1−x

2)3/2+C.



2√xdx = 1


So ∫e√x dx =

∫eu(2u) du = 2ueu − 2

∫eu du


= 2√xe√x − 2e

√x + C.

Trig substitution, or “reverse u-sub”How about ∫ √

1− x2 dx?

We could try letting u = 1− x2, so that x =√1− u. Further,

du = −2x dx = −2√1− u du. So∫ √

1− x2 dx = −12

∫ √u√

1− udu = −1

2

∫ √u

1− udu . . .


1− x2 dx?

We could try letting u = 1− x2, so that x =√1− u.

Further,du = −2x dx = −2

√1− u du. So∫ √

1− x2 dx = −12

∫ √u√

1− udu = −1

2

∫ √u

1− udu . . .


1− x2 dx?

We could try letting u = 1− x2, so that x =√1− u. Further,

du = −2x dx = −2√1− u du. So∫ √

1− x2 dx = −12

∫ √u√

1− udu = −1

2

∫ √u

1− udu . . .


1− x2 dx?

Instead of using a substitution that looks like u = f(x), we can trymaking a substitution that looks like x = f(u). Is there a functionf(u) such that 1− f2(u) is a perfect square? Think

cos2(u) + sin2(u) = 1.

Let x = cos(u). Then dx = − sin(u) du. So∫ √1− x2 dx =

∫ √1− cos2(u) · (− sin(u)) du

= −∫ √

sin2(u) · sin(u) du = −∫

sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?

Instead of using a substitution that looks like u = f(x), we can trymaking a substitution that looks like x = f(u).

Is there a functionf(u) such that 1− f2(u) is a perfect square? Think

cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?

Instead of using a substitution that looks like u = f(x), we can trymaking a substitution that looks like x = f(u). Is there a functionf(u) such that 1− f2(u) is a perfect square?

Think

cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.

Let x = cos(u).

Then dx = − sin(u) du. So∫ √1− x2 dx =

∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.

Let x = cos(u). Then dx = − sin(u) du.

So∫ √1− x2 dx =

∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √

sin2(u) · sin(u) du

= −∫

sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du

= 12(

12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 cos

−1(x) + C.


1− x2 dx?


cos2(u) + sin2(u) = 1.


∫ √1− cos2(u) · (− sin(u)) du

= −∫ √


sin2(u) du

= 12

∫cos(2u)− 1 du = 1

2(12 sin(2u)− u) + C

= 14 sin(2 cos

−1(x)))− 12 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

√1− x2

x

1

u

sin(2u) = 2 sin(u) cos(u) = 2√

1− x2 · x.

So∫ √1− x2 dx = · · · = −1

2 cos−1(x) + 1

4 sin(2 sin−1(x))) + C

= 12x√

1− x2 − 12 cos

−1(x) + C





√1− x2

x

1

u


1− x2 · x.

So∫ √1− x2 dx = · · · = −1

2 cos−1(x) + 1

4 sin(2 sin−1(x))) + C

= 12x√1− x2 − 1

2 cos−1(x) + C





√1− x2

x

1

u


1− x2 · x.

So∫ √1− x2 dx = · · · = −1

2 cos−1(x) + 1

4 sin(2 sin−1(x))) + C

= 12x√

1− x2 − 12 cos

−1(x) + C

Check against geometry:∫ √1− x2 dx = 1

2

(x√1− x2 − cos−1(x)

)+ C

Note: y =√1− x2 is the upper half of the graph of y2 + x2 = 1:

Recall: the area of a wedge with angle u of a circle or radius r is

A = (u/2π)πr2 = 12ur

2.

So, for example, the integral I =∫ 1a

√1− x2 dx should be

(area of the wedge)− (area of the triangle) = 12u−

12ah.

Since h =√1− a2 and u = cos−1(a), we have

I = 12 cos

−1(a)− 12a√1− a2.


2

(x√1− x2 − cos−1(x)

)+ C


u


A = (u/2π)πr2 = 12ur

2.




12ah.


I = 12 cos

−1(a)− 12a√1− a2.


2

(x√1− x2 − cos−1(x)

)+ C


a

1

u

h


A = (u/2π)πr2 = 12ur

2.




12ah.


I = 12 cos

−1(a)− 12a√1− a2.


2

(x√1− x2 − cos−1(x)

)+ C


a

1

u

h


A = (u/2π)πr2 = 12ur

2.




12ah.

Since h =√1− a2 and u = cos−1(a)

, we have

I = 12 cos

−1(a)− 12a√1− a2.


2

(x√1− x2 − cos−1(x)

)+ C


a

1

u

h


A = (u/2π)πr2 = 12ur

2.




12ah.


I = 12 cos

−1(a)− 12a√1− a2.


2

(x√1− x2 − cos−1(x)

)+ C

a

1

u

h

Geometrically,∫ 1

a

√1− x2 dx = 1

2u−12ah = 1

2 cos−1(a)− 1

2a√

1− a2.

Checking against the formula we computed:∫ 1

a

√1− x2 dx = 1

2

(x√1− x2 − cos−1(x)

) ∣∣1x=a

= 12

((1 · 0− cos−1(1))− (a

√1− a2 − cos−1(a))

)= 1

2 cos−1(a)− 1

2a√

1− a2 X (cos−1(1) = 0).


2

(x√1− x2 − cos−1(x)

)+ C

a

1

u

h

Geometrically,∫ 1

a

√1− x2 dx = 1

2u−12ah = 1

2 cos−1(a)− 1

2a√

1− a2.


a

√1− x2 dx = 1

2

(x√1− x2 − cos−1(x)

) ∣∣1x=a

= 12

((1 · 0− cos−1(1))− (a

√1− a2 − cos−1(a))

)

= 12 cos

−1(a)− 12a√

1− a2 X (cos−1(1) = 0).


2

(x√1− x2 − cos−1(x)

)+ C

a

1

u

h

Geometrically,∫ 1

a

√1− x2 dx = 1

2u−12ah = 1

2 cos−1(a)− 1

2a√

1− a2.


a

√1− x2 dx = 1

2

(x√1− x2 − cos−1(x)

) ∣∣1x=a

= 12

((1 · 0− cos−1(1))− (a

√1− a2 − cos−1(a))

)= 1

2 cos−1(a)− 1

2a√

1− a2 X (cos−1(1) = 0).

You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.

1.∫ √

1−x2

x2 dx using x = sin(u)

: dx = cos(u) du, so

I =∫ √1−sin2(u)

sin2(u)· cos(u) du =

∫ cos2(x)

sin2(u)du

=∫cot2(u) du = − cot(u)− u+ C

= − cot(sin−1(x))− sin−1(x) + C= −√1−x2

x − sin−1(x) + C

2.∫

1x2√1+x2

dx using x = tan(u)

: dx = sec2(u) du, so

I =∫

1

tan2(u)√

1+tan2(u)sec2(u) du =

∫ sec2(u)tan2(u) sec(u)

du

=∫ cos(u)

sin2(u)du = − sin−1(u) + C = − csc(tan−1(x)) + C

= −√1+x2

x + C.

3.∫

x√1+x2

dx two ways:

(a) Let x = tan(u) (b) Let u = 1 + x2.

=√1 + x2 + C


1.∫ √

1−x2

x2 dx using x = sin(u)

: dx = cos(u) du, so

I =∫ √1−sin2(u)


∫ cos2(x)

sin2(u)du


= − cot(sin−1(x))− sin−1(x) + C

= −√1−x2

x − sin−1(x) + C

2.∫

1x2√1+x2

dx using x = tan(u)

: dx = sec2(u) du, so

I =∫

1

tan2(u)√



du

=∫ cos(u)


= −√1+x2

x + C.

3.∫

x√1+x2

dx two ways:


=√1 + x2 + C


1.∫ √

1−x2

x2 dx using x = sin(u): dx = cos(u) du, so

I =∫ √1−sin2(u)


∫ cos2(x)

sin2(u)du


= − cot(sin−1(x))− sin−1(x) + C= −√1−x2

x − sin−1(x) + C

2.∫

1x2√1+x2

dx using x = tan(u): dx = sec2(u) du, so

I =∫

1

tan2(u)√



du

=∫ cos(u)


= −√1+x2

x + C.

3.∫

x√1+x2

dx two ways:


=√1 + x2 + C


3.∫

x√1+x2

dx two ways:

(a) Let x = tan(u):

dx = sec2(u) du, so that∫x√

1 + x2dx =

∫tan(u)√

1 + tan2(x)sec2(u) du

=

∫sec(u) tan(u) du

= sec(u) + C =√1 + x2 + C

x

1

√1 + x2

u

(b) Let u = 1 + x2:

Then du = 2x dx, so that∫x√

1 + x2dx = 1

2

∫u−1/2 du = 1

2 ·2u1/2+C =

√1 + x2+C.


3.∫

x√1+x2

dx two ways:

(a) Let x = tan(u): dx = sec2(u) du, so that∫x√

1 + x2dx =

∫tan(u)√


=

∫sec(u) tan(u) du

= sec(u) + C =√1 + x2 + C

x

1

√1 + x2

u

(b) Let u = 1 + x2:


1 + x2dx = 1

2

∫u−1/2 du = 1

2 ·2u1/2+C =

√1 + x2+C.


3.∫

x√1+x2

dx two ways:


1 + x2dx =

∫tan(u)√


=

∫sec(u) tan(u) du = sec(u) + C

=√1 + x2 + C

x

1

√1 + x2

u

(b) Let u = 1 + x2:


1 + x2dx = 1

2

∫u−1/2 du = 1

2 ·2u1/2+C =

√1 + x2+C.


3.∫

x√1+x2

dx two ways:


1 + x2dx =

∫tan(u)√


=

∫sec(u) tan(u) du = sec(u) + C =

√1 + x2 + C

x

1

√1 + x2

u

(b) Let u = 1 + x2:


1 + x2dx = 1

2

∫u−1/2 du = 1

2 ·2u1/2+C =

√1 + x2+C.


3.∫

x√1+x2

dx two ways:


1 + x2dx =

∫tan(u)√


=

∫sec(u) tan(u) du = sec(u) + C =

√1 + x2 + C

x

1

√1 + x2

u

(b) Let u = 1 + x2: Then du = 2x dx, so that∫x√

1 + x2dx = 1

2

∫u−1/2 du = 1

2 ·2u1/2+C =

√1 + x2+C.

Another geometric exampleCompute the area of an ellipse with minor radius 3 and majorradius 4.

x2

42+ y2

32= 1

The desired area is half the area under the curvey = 3

√1− (x/4)2. So A = 2 · 3

∫ 4−4√

1− (x/4)2 dx.Let x/4 = sin(u), so that dx = 4 cos(u) du. Thus,∫ √

1− x2/42 dx =

∫ √1− sin2(u) 4 cos(u) du = 4

∫cos2(u) du

= 4·12(u+cos(u) sin(u))+C = 2(sin−1(x/4) + (x/4)

√1− (x/4)2

)+C


x2

42+ y2

32= 1


√1− (x/4)2.

So A = 2 · 3∫ 4−4√

1− (x/4)2 dx.Let x/4 = sin(u), so that dx = 4 cos(u) du. Thus,∫ √

1− x2/42 dx =

∫ √1− sin2(u) 4 cos(u) du = 4

∫cos2(u) du


√1− (x/4)2

)+C


x2

42+ y2

32= 1


√1− (x/4)2. So A = 2 · 3

∫ 4−4√1− (x/4)2 dx.

Let x/4 = sin(u), so that dx = 4 cos(u) du. Thus,∫ √1− x2/42 dx =

∫ √1− sin2(u) 4 cos(u) du = 4

∫cos2(u) du


√1− (x/4)2

)+C


x2

42+ y2

32= 1


√1− (x/4)2. So A = 2 · 3

∫ 4−4√1− (x/4)2 dx.

Let x/4 = sin(u), so that dx = 4 cos(u) du.

Thus,∫ √1− x2/42 dx =

∫ √1− sin2(u) 4 cos(u) du = 4

∫cos2(u) du


√1− (x/4)2

)+C


x2

42+ y2

32= 1


√1− (x/4)2. So A = 2 · 3

∫ 4−4√1− (x/4)2 dx.


∫ √1− sin2(u) 4 cos(u) du

= 4

∫cos2(u) du


√1− (x/4)2

)+C


x2

42+ y2

32= 1


√1− (x/4)2. So A = 2 · 3

∫ 4−4√1− (x/4)2 dx.


∫ √1− sin2(u) 4 cos(u) du = 4

∫cos2(u) du


√1− (x/4)2

)+C


x2

42+ y2

32= 1


√1− (x/4)2. So

A = 6

∫ 4

−4

√1− (x/4)2 dx

= 6(2(sin−1(x/4) + (x/4)

√1− (x/4)2

)) ∣∣4x=−4

= 12((sin−1(1) +

√1− 1

)−(sin−1(x− 1) + (−1)

√1− 1

))= 12(π/2− (−π/2)) = 12π .


x2

42+ y2

32= 1


√1− (x/4)2. So

A = 6

∫ 4

−4

√1− (x/4)2 dx

= 6(2(sin−1(x/4) + (x/4)

√1− (x/4)2

)) ∣∣4x=−4

= 12((sin−1(1) +

√1− 1

)−(sin−1(x− 1) + (−1)

√1− 1

))

= 12(π/2− (−π/2)) = 12π .


x2

42+ y2

32= 1


√1− (x/4)2. So

A = 6

∫ 4

−4

√1− (x/4)2 dx

= 6(2(sin−1(x/4) + (x/4)

√1− (x/4)2

)) ∣∣4x=−4

= 12((sin−1(1) +

√1− 1

)−(sin−1(x− 1) + (−1)

√1− 1

))= 12(π/2− (−π/2))

= 12π .


x2

42+ y2

32= 1


√1− (x/4)2. So

A = 6

∫ 4

−4

√1− (x/4)2 dx

= 6(2(sin−1(x/4) + (x/4)

√1− (x/4)2

)) ∣∣4x=−4

= 12((sin−1(1) +

√1− 1

)−(sin−1(x− 1) + (−1)

√1− 1

))= 12(π/2− (−π/2)) = 12π .

Completing the squareCompute ∫

1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4 = 4((x−22

)2 − 1).

Recall, this is called completing the square. In general, if you’reinterested 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 − (b/2a)2

)+c

=(√a(x+ b/2a)2

)− 1

a(b/2)2 + c.


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4 = 4((x−22

)2 − 1).

Recall, this is called completing the square.

In general, if you’reinterested in rewriting


So


)+c = a

((x+ b/2a)2 − (b/2a)2

)+c

=(√a(x+ b/2a)2

)− 1

a(b/2)2 + c.


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4 = 4((x−22

)2 − 1).



So


)+c = a

((x+ b/2a)2 − (b/2a)2

)+c

=(√a(x+ b/2a)2

)− 1

a(b/2)2 + c.


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4 = 4((x−22

)2 − 1).



So


)+c

= a((x+ b/2a)2 − (b/2a)2

)+c

=(√a(x+ b/2a)2

)− 1

a(b/2)2 + c.


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4 = 4((x−22

)2 − 1).



So


)+c = a

((x+ b/2a)2 − (b/2a)2

)+c

=(√a(x+ b/2a)2

)− 1

a(b/2)2 + c.


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.

Let (x− 2)/2 = sec(u). Then 12 dx = sec(u) tan(u) du, so that

I =

∫1

2√

sec2(u)− 12 sec(u) tan(u) du

=

∫sec(u) tan(u)

tan(u)du = ln | sec(u) + tan(u)|+ C

√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx

=

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.

Let (x− 2)/2 = sec(u).

Then 12 dx = sec(u) tan(u) du, so that

I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.

Let (x− 2)/2 = sec(u). Then 12 dx = sec(u) tan(u) du

, so that

I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)

tan(u)du

= ln | sec(u) + tan(u)|+ C

√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u

= ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C


1√x2 − 4x

dx

Rewrite

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

Then∫1√

x2 − 4xdx =

∫1√

(x− 2)2 − 4dx =

∫1

2√

((x− 2)/2)2 − 1dx.


I =

∫1

2√


=

∫sec(u) tan(u)


√(x−22

)2 − 1

1

x−22

u = ln

∣∣∣∣x−22 +√(x−22 )2 − 1

∣∣∣∣+ C

You try:

Calculate the following integrals. Be sure to simplify your answers.Remember, the Pythagorean identities are

cos2(u) + sin2(u) = 1 and 1 + tan2(u) = sec2(u).

1.∫

1√x2−1 dx

(Hint: Let x = sec(u).)

Ans: ln(x+√x2 − 1) + C.

2.∫ √

3− x2 dx

(Hint: Let x =√3 sin(u).)

Ans: 12

(√3− x2 + 3 sin−1(x/

√3))+ C.

3.∫

ex

e2x√1+e2x

dx

(Hint: Let ex = tan(u).)

Ans: −e−x√1 + e2x + C.

You try:

Calculate the following integrals. Be sure to simplify your answers.Remember, the Pythagorean identities are

cos2(u) + sin2(u) = 1 and 1 + tan2(u) = sec2(u).

1.∫

1√x2−1 dx (Hint: Let x = sec(u).)

Ans: ln(x+√x2 − 1) + C.

2.∫ √

3− x2 dx (Hint: Let x =√3 sin(u).)

Ans: 12

(√3− x2 + 3 sin−1(x/

√3))+ C.

3.∫

ex

e2x√1+e2x

dx (Hint: Let ex = tan(u).)

Ans: −e−x√1 + e2x + C.

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