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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
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)
∫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
=
∫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)
∫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
=
∫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)
∫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
=
∫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)
∫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
=
∫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)
∫cos2(x) sin2(x) dx =
∫(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
=
∫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)
∫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
=
∫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)
∫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
= 132(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)
∫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
=
∫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)
∫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 = 1
2
∫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)
∫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 = 1
2
∫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)
∫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 = 1
2
∫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)
∫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 = 1
2
∫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)
∫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 = 1
2
∫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.
(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
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
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
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
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
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
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 .
(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
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
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
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
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
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
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
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
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
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 .
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.
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.
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
= −1223u
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.
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.
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.
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.
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
= 12udx, 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.
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.
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.
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.
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.
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 . . .
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 . . .
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 . . .
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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
= 12(
12 sin(2u)− u) + C
= 14 sin(2 cos
−1(x)))− 12 cos
−1(x) + C.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
Trig substitution, or “reverse u-sub”How about ∫ √
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.
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
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
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 − 1
2 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 − 1
2 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
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.
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.
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:
u
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.
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:
a
1
u
h
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.
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:
a
1
u
h
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.
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:
a
1
u
h
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.
Check against geometry:∫ √1− x2 dx = 1
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).
Check against geometry:∫ √1− x2 dx = 1
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).
Check against geometry:∫ √1− x2 dx = 1
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))
)
= 12 cos
−1(a)− 12a√
1− a2 X (cos−1(1) = 0).
Check against geometry:∫ √1− x2 dx = 1
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).
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
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
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
You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.
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.
You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.
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.
You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.
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.
You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.
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.
You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.
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.
You try:Calculate the following integrals using the suggested substitution.Be sure to simplify your answers.
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.
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
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
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
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
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
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
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
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
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
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 = 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 .
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 = 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 .
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 = 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 .
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 = 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.
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.
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.
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.
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.
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.
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
Completing the squareCompute ∫
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
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.
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.