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Section 6.2 Trigonometric Integrals and Substitutions
2010 Kiryl Tsishchanka
Trigonometric Integrals and Substitutions
This section consists of two parts:
Part I: Trigonometric Integrals. We will distinguish three main cases:
Case A: Integrals of type
Z
sinm x cosn xdx
where m and n are nonnegative integers.
METHOD OF INTEGRATION:
(i) If m is odd, then u = cos x.
(ii) If n is odd, then u = sin x.
(iii) If both m and n are even, then use the identities
1
sin2 x = (1 − cos 2x),
2
1
cos2 x = (1 + cos 2x)
2
or(and) sometimes
sin x cos x =
EXAMPLE 1: Find
Z
1
sin 2x
2
sin3 xdx.
Solution: We have
Z
3
sin xdx =
=−
Z
EXAMPLE 2: Find
Z
2
sin x · sin xdx =
Z


 d cos x = du
(1 − cos x) sin xdx = 
 − sin xdx = du

2
1
1
(1 − u2 )du = −u + u3 + C = − cos x + cos3 x + C
3
3
Z
cos x = u
sin5 xdx.
1
sin xdx = −du






Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 2: Find
Z
2010 Kiryl Tsishchanka
sin5 xdx.
Solution: We have
Z
Z
Z
Z
5
4
2
2
sin xdx = sin x · sin xdx = (sin x) sin xdx = (1 − cos2 x)2 sin xdx
cos x = u


 d cos x = du
=
 − sin xdx = du



Z
Z

 = − (1 − u2 )2 du = − (1 − 2u2 + u4 )du


sin xdx = −du
2
1
2
1
= −u + u3 − u5 + C = − cos x + cos3 x − cos5 x + C
3
5
3
5
EXAMPLE 3: Find
Z
sin2 x cos3 xdx.
Solution: We have
Z
Z
Z
2
3
2
2
sin x cos xdx = sin x cos x · cos xdx = sin2 x(1 − sin2 x) cos xdx

sin x = u



=  d sin x = du  =
cos xdx = du
EXAMPLE 4: Find
Z √
3
Z
2
2
u (1 − u )du =
Z
1
1
1
1
(u2 − u4 )du = u3 − u5 + C = sin3 x − sin5 x + C
3
5
3
5
sin x cos5 xdx.
2
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 4: Find
Z √
3
2010 Kiryl Tsishchanka
sin x cos5 xdx.
Solution: We have
Z √
Z √
Z √
3
3
3
5
4
sin x cos xdx =
sin x cos x · cos xdx =
sin x(cos2 x)2 · cos xdx


sin x = u
Z √
Z
√


3
2
2
3
=
sin x(1 − sin x) · cos xdx =  d sin x = du  =
u(1 − u2 )2 du
cos xdx = du
=
Z
u
=
Z
(u1/3 − 2u7/3 + u13/3 )du =
1/3
2
4
(1 − 2u + u )du =
Z
=
(u1/3 − 2u2+1/3 + u4+1/3 )du
u7/3+1
u13/3+1
u1/3+1
−2
+
+C
1/3 + 1
7/3 + 1 13/3 + 1
u4/3
u10/3 u16/3
−2
+
+C
4/3
10/3 16/3
3
3
3
= u4/3 − u10/3 + u16/3 + C
4
5
16
=
EXAMPLE 5: Find
Z
3
3
3 4/3
sin x − sin10/3 x +
sin16/3 x + C
4
5
16
sin2 x cos2 xdx.
Solution: We have
2
Z
Z
Z Z
Z
1
1
1
1
2
2
2
2
sin x cos xdx = (sin x cos x) dx =
sin 2xdx =
sin 2x dx =
(1 − cos 4x)dx
2
4
4
2


4x = u


 d(4x) = du 
Z
Z


1
1
1
1


(1 − cos 4x)dx =  4dx = du  =
(1 − cos u)du = (u − sin u) + C = (4x − sin 4x) + C
=
8
32
32
 32



1
dx = du
4
EXAMPLE 6: Find
Z
cos4 xdx.
3
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 6: Find
Z
2010 Kiryl Tsishchanka
cos4 xdx.
Solution: We have
2
Z
Z
Z Z
1 + cos 2x
1
4
2
2
cos xdx = (cos x) dx =
(1 + cos 2x)2 dx
dx =
2
4
Z
Z 1
1
1
2
=
(1 + 2 cos 2x + cos 2x)dx =
1 + 2 cos 2x + (1 + cos 4x) dx
4
4
2
Z 1
3
1
1 3
1
1
1 1 1
=
+ 2 cos 2x + cos 4x dx = · x + · 2 · sin 2x + · · sin 4x + C
4
2
2
4 2
4
2
4 2 4
1
1
3
sin 4x + C
= x + sin 2x +
8
4
32
Case B: Integrals of type
Z
tanm x secn xdx
Midterms where m and n are nonnegative integers.
METHOD OF INTEGRATION:
(i) If m is odd, then u = sec x.
(ii) If n is even, then u = tan x.
(iii) In other cases the guidelines are not as clear-cut.
EXAMPLE 7: Find
Z
tan xdx.
Solution 1: We have
Z
tan xdx =
Z
sin x
dx =
cos x
Z
cos x = u


 d cos x = du
1
· sin xdx = 
 − sin xdx = du
cos x

sin xdx = −du


Z

 = − 1 du

u

= − ln |u| + C = − ln | cos x| + C = ln | cos x|−1 + C = ln | sec x| + C
Solution 2: We have
Z
tan xdx =
Z

tan x sec x

dx = 
sec x
sec x = u
d sec x = du
sec x tan xdx = du
4


=
Z
1
du = ln |u| + C = ln | sec x| + C
u
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 8: Find
Z
(a) csc xdx
2010 Kiryl Tsishchanka
Z
(b)
sec xdx
Solution 1:
(a) We have

x
=u
2



 x
 Z
 d
Z
Z
Z
Z
Z
=
du


dx
dx
2du
du
2du
2


x = 
csc xdx =
=
=
=
=

sin x
sin (2u)
2 sin u cos u
sin u cos u

 1
sin 2 ·

dx = du 
2

 2
dx = 2du


tan u = v
Z
Z
Z
Z
Z
sec2 udu
sec2 udu
sec2 udu 
dv
sec2 udu

d
tan
u
=
dv
=
=
=
=
=
=


sin
u
cos
u
sin
u
sec2 u sin u cos u
tan u
v
cos2 u
cos u
sec2 udu = dv
x = ln |v| + C = ln | tan u| + C = ln tan
+C
2
In short,
2 x
Z
Z
Z
Z
Z
sec
dx
dx
dx
dx
2
x
csc xdx =
=
x =
x
x =
x
x
sin x
2
cos
sin
cos
sin 2 ·
2 sin
2 sec
2
2
2
2
2
2
x


=v
tan
2


x
x


Z sec2
x = dv  Z dv
dx  d tan


2
2
=
= ln |v| + C = ln tan
=
+C
x =
1
x
 sec2
v
2
2 tan
dx = dv 


2
2

 2 x
2
dx = 2dv
sec
2
This can be rewritten as ln |csc x − cot x| + C (see Appendix II).

(b) We have
Z
sec xdx =
Z
dx
=
cos x
Z




=

π

sin x +
2
dx
π
x+ =u
2
π
= du
d x+
2
dx = du

 Z

du
=

sin u

u π
which is ln tan
+ C by (a). Substituting x + for u, we get
2
2


π
Z
x+
2
 + C = ln tan x + π + C
sec xdx = ln tan 
2
2 4
This can be rewritten as ln |sec x + tan x| + C (see Appendix II).
5
Section 6.2 Trigonometric Integrals and Substitutions
2010 Kiryl Tsishchanka
Solution 2:
(a) We have
Z
csc xdx =
1
dx =
sin x
Z
=−
Z
sin x
dx =
sin2 x
Z
1
du = −
(1 − u)(1 + u)
Z
Z
cos x = u


 d cos x = du
sin x

dx
=
 − sin xdx = du
1 − cos2 x

1
2


Z

1
=−
du

1 − u2

sin xdx = −du
Z
Z
1
1
1
1
1
1
du = −
+
du −
du
1−u 1+u
2
1−u
2
1+u
1
1
1
1 1 − u 1 − cos x
= ln |1 − u| − ln |1 + u| + C = ln + C = ln
+C
2
2
2
1 + u
2
1 + cos x
This can be rewritten as ln |csc x − cot x| + C (see Appendix II).
(b) We have
Z
cos x
dx =
cos2 x
Z
1
2
sec xdx =
Z
1
dx =
cos x
=
Z
1
du =
(1 − u)(1 + u)
Z
Z
sin x = u


cos x


2 dx =  d sin x = du  =
1 − sin x
cos xdx = du
1
1
+
1−u 1+u
1
du =
2
Z
Z
1
1
du +
1−u
2
1
du
1 − u2
Z
1
du
1+u
1
1
1 1 + u 1 + sin x
1
+ C = ln
+C
= − ln |1 − u| + ln |1 + u| + C = ln 2
2
2
1 − u
2
1 − sin x
This can be rewritten as ln |sec x + tan x| + C (see Appendix II).
Solution 3:
(a) We have
Z
csc xdx =
=
Z
Z
csc x
csc x − cot x
dx =
csc x − cot x
Z
csc x − cot x = u

csc2 x − csc x cot x

dx = 
csc x − cot x
d(csc x − cot x) = du
(− csc x cot x + csc2 x)dx = du
1
du = ln |u| + C = ln | csc x − cot x| + C
u



(b) We have
Z
sec xdx =
Z
sec x
sec x + tan x
dx =
sec x + tan x
Z

2
sec x + sec x tan x

dx = 
sec x + tan x
1
du = ln |u| + C = ln | sec x + tan x| + C
u
Z
EXAMPLE 9: Find tan8 x sec4 xdx.
=
Z
6
sec x + tan x = u
d(sec x + tan x) = du
(sec x tan x + sec2 x)dx = du



Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 9: Find
Solution: We have
Z
Z
tan8 x sec4 xdx =
=
Z
tan8 x sec4 xdx.
Z
u8 (1 + u2)du =
EXAMPLE 10: Find
Z
2010 Kiryl Tsishchanka
Z
tan8 x sec2 x · sec2 xdx =
Z
tan x = u




tan8 x(1 + tan2 x) sec2 xdx =  d tan x = du 
sec2 xdx = du
1
1
1
1
(u8 + u10 )du = u9 + u11 + C = tan9 x +
tan11 x + C
9
11
9
11
tan3 xdx.
Solution 1: We have
Z
Z
Z
3
2
tan xdx = tan x · tan xdx = (sec2 x − 1) · tan xdx


sec x = u
Z
Z
(u2 − 1)du
(sec2 x − 1) · sec x tan xdx 

d sec x = du
=
=
=

sec x
u
sec x tan xdx = du
Z 1
u2
1
=
u−
du =
− ln |u| + C = sec2 x − ln | sec x| + C
u
2
2
Solution 2: We have
Z
Z
Z
Z
3
2
2
tan xdx = tan x · tan xdx = (sec x − 1) · tan xdx = (sec2 x tan x − tan x)dx
=
To evaluate
Z
Z
Z
sec x tan xdx −
Z
tan xdx = [by Example 7] =
Z
sec2 x tan xdx − ln | sec x|
Z
udu =
sec2 x tan xdx, we either
2
sec x tan xdx =
Z
or
Z
Therefore
2

sec x · sec x tan xdx = 
2
sec x = u



d sec x = du
sec x tan xdx = du
tan x = u



sec x tan xdx =  d tan x = du  =
sec2 xdx = du
Z
udu =

=
1
u2
+ C = sec2 x + C
2
2
1
u2
+ C = tan2 x + C
2
2
1
sec2 x − ln | sec x| + C
2
or
1
tan2 x − ln | sec x| + C
2
Other possible answers are
Z
1
tan3 xdx = sec2 x + ln | cos x| + C
2
or
1
tan2 x + ln | cos x| + C
2
Z
tan3 xdx =
7
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 11: Find
Z
2010 Kiryl Tsishchanka
sec3 xdx.
Solution: We have
Z
3
sec xdx =
Z
2


sec x · sec xdx = 
sec x = u
sec2 xdx = dv
d(sec x) = du
tan x = v
It follows that
2
so
1
sec xdx =
2
3

 = sec x tan x −
Z
sec x tan2 xdx
sec x tan xdx = du
Z
Z
2
= sec x tan x − sec x(sec x − 1)dx = sec x tan x − (sec3 x − sec x)dx
= sec x tan x −
Z

Z
Z
3
sec xdx +
3
Z
sec xdx = sec x tan x +
sec xdx
Z
sec xdx
Z
1
sec x tan x + sec xdx = [by Example 8] = (sec x tan x + ln | sec x + tan x|) + C
2
Case C: Integrals of type
Z
sin αx sin βxdx,
Z
sin αx cos βxdx or
Z
cos αx cos βxdx
METHOD OF INTEGRATION: Use the identities
1
sin A sin B = [cos(A − B) − cos(A + B)]
2
1
cos A cos B = [cos(A + B) + cos(A − B)]
2
1
sin A cos B = [sin(A + B) + sin(A − B)]
2
EXAMPLE 12: Find
Z
cos 3x cos 2xdx.
8
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 12: Find
Z
Solution 1: We have
Z
2010 Kiryl Tsishchanka
cos 3x cos 2xdx.
cos 3x cos 2xdx =
1
cos A cos B = [cos(A + B) + cos(A − B)]
2
1
[cos(3x + 2x) + cos(3x − 2x)]dx
2
Z
Z
1
1
cos 5xdx +
cos xdx
=
2
2
=
Z
=
1
1
sin 5x + sin x + C
10
2
Solution 2: We have
Z

cos 3x = u

cos 2xdx = dv
1

sin 2x = v 
2

cos 3x cos 2xdx =  d(cos 3x) = du
−3 sin 3xdx = du
1
= cos 3x · sin 2x −
2
3
1
= cos 3x sin 2x +
2
2

1
sin 2x · (−3) sin 3xdx
2
Z
Z
sin 2x sin 3xdx

sin 2xdx = dv
1

− cos 2x = v 
2
sin 3x = u

=  d(sin 3x) = du
3 cos 3xdx = du
Z 1
1
1
3
= cos 3x sin 2x +
sin 3x · −
cos 2x −
−
cos 2x · 3 cos 3xdx
2
2
2
2
1
3
9
= cos 3x sin 2x − sin 3x cos 2x +
2
4
4
therefore
5
−
4
hence
Z
Z
Z
cos 2x cos 3xdx
cos 3x cos 2xdx =
1
3
cos 3x sin 2x − sin 3x cos 2x + C1
2
4
1
3
cos 3x sin 2x − sin 3x cos 2x + C1
2
4
4
cos 3x cos 2xdx = −
5
3
2
= − cos 3x sin 2x + sin 3x cos 2x + C
5
5
9
Section 6.2 Trigonometric Integrals and Substitutions
2010 Kiryl Tsishchanka
Part II: Trigonometric Substitutions. Here we deal with integrals that involve
√
√
√
a2 − x2 ,
a2 + x2 ,
x2 − a2
METHOD OF INTEGRATION:
√
π
π
(i) If a2 − x2 , then x = a sin θ, − ≤ θ ≤ .
2
2
√
π
π
(ii) If a2 + x2 , then x = a tan θ, − < θ < .
2
2
√
3π
π
(iii) If x2 − a2 , then x = a sec θ, 0 ≤ θ < or π ≤ θ <
.
2
2
Z √
EXAMPLE 13: Find
1 − x2 dx.
Solution 1: We have
Z √



1 − x2 dx = 


Z √
Solution 2: We have
 √
1−
dx = d(sin θ)







p
√
sin 2θ = 2 sin θ cos θ = 2x 1 − x2
Therefore
x2 dx
x = sin θ
dx = cos θdθ
√
√
1 − x2 = 1 − sin2 θ = cos2 θ = | cos θ| = cos θ
Z
Z
Z
1
1
1
2
(1 + cos 2θ)dθ = θ + sin 2θ + C
= cos θ · cos θdθ = cos θdθ =
2
2
4
Note that
Z √

1 − x2 = u
1 − x2 dx =
 √
2

=  d( 1 − x ) = du

−x
√
dx = du
1 − x2
dx = dv
1 √
1 −1
sin x + x 1 − x2 + C
2
2

Z

x = v  = √1 − x2 · x − x · √ −x dx


1 − x2
Z
Z √
√
1 − x2 − 1
1 − x2
1
−x2
2
2
√
√
√
dx = x 1 − x −
dx = x 1 − x −
−√
dx
1 − x2
1 − x2
1 − x2
1 − x2
Z √
Z √
Z
√
√
1
1 − x2 dx + √
1 − x2 dx + sin−1 x
dx = x 1 − x2 −
= x 1 − x2 −
2
1−x
Therefore
Z √
√
1 − x2 dx = x 1 − x2 + sin−1 x + C1
2
√
= x 1 − x2 −
so
Z
Z √
1 √
1
1 − x2 dx = x 1 − x2 + sin−1 x + C
2
2
10
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 14: Find
Z √
2010 Kiryl Tsishchanka
1 + x2 dx.
Solution: We have
Z √
x = tan θ


dx = d(tan θ)


1 + x2 dx = 







dx = sec2 θdθ
√
√
√
1 + x2 = 1 + tan2 θ = sec2 θ = | sec θ| = sec θ
Z
Z
1
2
= sec θ · sec θdθ = sec3 θdθ = [Example 11] = (sec θ tan θ + ln | sec θ + tan θ|) + C
2
√
1 √
= (x 1 + x2 + ln(x + 1 + x2 )) + C
2
EXAMPLE 15: Find
Z √
x2 − 1dx.
Solution: We have
Z √
x2
x = sec θ



− 1dx = 


dx = d(sec θ)
dx = sec θ tan θdθ
√
x2 − 1 = sec2 θ − 1 = tan2 θ = | tan θ| = tan θ
Z
Z
= tan θ · sec θ tan θdθ = tan2 θ sec θdθ
=
=
Z
Z
√
√
2
(sec θ − 1) sec θdθ =
3
sec θdθ −
Z
Z






(sec3 θ − sec θ)dθ
sec θdθ = [Examples 8 and 11]
1
= (sec θ tan θ + ln | sec θ + tan θ|) − ln | sec θ + tan θ| + C
2
=
1
1
sec θ tan θ + ln | sec θ + tan θ| − ln | sec θ + tan θ| + C
2
2
=
1
1
sec θ tan θ − ln | sec θ + tan θ| + C
2
2
1
= (sec θ tan θ − ln | sec θ + tan θ|) + C
2
√
1 √
= (x x2 − 1 − ln |x + x2 − 1|) + C
2
11
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 16: Find
Z √
2010 Kiryl Tsishchanka
4 − x2
dx.
x2
Solution 1: We have
Z √
4−
x2
x2




dx = 



x = 2 sin θ
dx = d(2 sin θ)
dx = 2 cos θdθ
p
p
√
4 − x2 = 4 − 4 sin2 θ = 4(1 − sin2 θ) = 4 cos2 θ = 2| cos θ| = 2 cos θ
Z
Z
Z
Z 2 cos θ
sin2 θ
cos2 θ
1 − sin2 θ
1
=
· 2 cos θdθ =
dθ =
dθ =
−
dθ
4 sin2 θ
sin2 θ
sin2 θ
sin2 θ sin2 θ
√
=
Note that
cot θ =
Therefore
Z √
Solution 2: We have

√
√
√
Z






(csc2 θ − 1)dθ = − cot θ − θ + C
4 − x2
x
4 − x2
4 − x2
−1 x
+C
dx
=
−
−
sin
x2
x
2
!!
2!!!
!
!!
!
θ
!
√
x
4 − x2

1
dx = dv 
4− =u

x2
Z Z √
 √

1
4 − x2
1
−x
1

 √
2
2
√
−
−
dx =  d( 4 − x ) = du − = v  = 4 − x · −
dx
2


x
x2
x
x
4
−
x


−x
√
dx = du
2
4−x
√
√
Z
x
4 − x2
4 − x2
dx
+C
− √
=−
− sin−1
=−
x
x
2
4 − x2
EXAMPLE 17: Find
Z
x2
x2
√
1
dx.
x2 − 9
12
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 17: Find
Z
x2
√
2010 Kiryl Tsishchanka
1
dx.
x2 − 9
Solution 1: We have
Z


1
√
dx = 

x2 x2 − 9

Note that
Therefore
x = 3 sec θ

dx = d(3 sec θ)






dx = 3 sec θ tan θdθ
√
p
x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3| tan θ| = 3 tan θ
Z
Z
1
1
1
=
cos θdθ = sin θ + C
· 3 sec θ tan θdθ =
2
9 sec θ · 3 tan θ
9
9
!
x !!! √
√
!
x2 − 9
x2 − 9
!!
!
sin θ =
!! θ
x
3
√
Z
2
x −9
1
1
√
+C
dx = ·
2
2
9
x
x x −9
√
√
Solution 2: We first note that
√
Z √ 2
Z √ 2
Z
x −9
x −9
x2 − 9
1
1
√
dx =
−
dx
dx =
x2 (x2 − 9)
9
x2 − 9
x2
x2 x2 − 9
√
Z x2 − 9
1
1
√
dx
−
=
9
x2
x2 − 9
Z
Z √ 2
1
x −9
1
1
√
=
dx −
dx
2
9
9
x2
x −9
We have
 √

1
dx
=
dv
x2 − 9 = u


x2
Z √ 2
Z √


√
x −9
1
1
2 −9·
 d( x2 − 9) = du − = v 
dx
=
dx
=
x


x2
x2
x


x
√
dx = du
x2 − 9
Z √
1
x
1
√
−
−
dx
= x2 − 9 −
2
x
x
x −9
√
Z
x2 − 9
1
=−
+ √
dx
2
x
x −9
Therefore
Z
Z √ 2
Z
x −9
1
1
1
1
√
√
dx
dx =
dx −
2
9
9
x2
x x2 − 9
x2 − 9
√
√
Z
Z
x2 − 9 1
x2 − 9
1
1
1
1
1
√
√
dx + ·
dx = ·
−
+C
=
9
9
x
9
9
x
x2 − 9
x2 − 9
13
Section 6.2 Trigonometric Integrals and Substitutions
Solution 3 (version 1): Note that
Z
2010 Kiryl Tsishchanka
1
√
dx can be rewritten as
2
x x2 − 9
Z
xm (a + bxn )p dx with
1
m = −2, n = 2, p = − , a = −9, and b = 1
2
It is known that if
m+1
+ p is an integer (which is exactly the case)
n
then
b+
a
= uN , where N is the denominator of p
xn
Therefore we have






Z

1

√
dx = 
2
2

x x −9





1
=
9
9
1 − 2 = u2 =⇒
x
9
d 1 − 2 = du2
x
r
9
1 − 2 = u =⇒
x
√
x2 − 9
=u
x
18
dx = 2udu
x3
1
1
x
1
1 1
1
1
√
· dx = du =⇒ 3 · √
dx = du =⇒
dx = du
3
x u
9
x
9
9
x2 − 9
x2 x2 − 9
Z
1
1
du = u + C = ·
9
9
Solution 3 (version 2): We have
 √
√
x2 − 9
+C
x
x2 − 9
=u
x
√ 2
x −9
= du
d
x





Z

1

√
dx = 
2
2

x x −9
 √9
dx = du
 2 2
 x x −9


1
1
√
dx = du
9
x2 x2 − 9






√
 1Z
x2 − 9
1
1

du = u + C = ·
+C
=
 9
9
9
x





14















Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 18: Find
√
3Z 3/2
2010 Kiryl Tsishchanka
x3
dx.
(4x2 + 9)3/2
0
Solution: We have






√

3Z 3/2

3

x

dx
=

(4x2 + 9)3/2

0






Since
√
3
x = tan θ
2
3
dx = d
tan θ
2
3
dx = sec2 θdθ
2
s r
2
√
√
3
9
4x2 + 9 = 4
tan θ + 9 = 4 · tan2 θ + 9 = 9 tan2 θ + 9
2
4
p
√
= 9(tan2 θ + 1) = 9 sec2 θ = 3| sec θ| = 3 sec θ


















4x2 + 9 = 3 sec θ, we get
√
(4x2 + 9)3/2 = ( 4x2 + 9)3 = (3 sec θ)3 = 27 sec3 θ
We also note that since x =
3
tan θ, we have
2
if x = 0, then θ = 0
√
if x = 3 3/2, then θ = π/3
and
Therefore
√
3Z 3/2
x3
dx =
(4x2 + 9)3/2
0
3
=
16
tan3 θ 3
3
8
· sec2 θdθ =
3
27 sec θ 2
16
0
Zπ/3
0
3
=
16
Zπ/3 27
Zπ/3
tan3 θ
3
dθ =
sec θ
16
0

cos θ = u



1 − cos2 θ
d(cos θ) = du  = − 3
sin θdθ = 


2
cos θ
16
− sin θdθ = du
EXAMPLE 19: Find
Z
√
x
dx.
x2 − 9
15
sin3 θ
3
dθ =
2
cos θ
16
0
Z1/2
1 − u2
3
du = −
2
u
16
1
1/2
Z1/2
3
1
1
3
3
−2
=
u+
+ 2 − (1 + 1) =
(1 − u )du =
16
u 1
16
2
32
1
Zπ/3
Zπ/3
0
Z1/2
(u−2 − 1)du
1
sin2 θ
· sin θdθ
cos2 θ
Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE 19: Find
Z
√
2010 Kiryl Tsishchanka
x
dx.
−9
x2
Solution 1: We have
Z

x2 − 9 = u

 d(x2 − 9) = du

x
√
dx = 
 2xdx = du
x2 − 9


1
xdx = du
2



Z
Z
 1
du
1 u−1/2+1
1
−1/2
=
√
u
du
=
=
·
+C
 2
u
2
2 −1/2 + 1


=
√
u+C =
√
x2 − 9 + C
Solution 2: We have
Z


x
√
dx = 

x2 − 9

dx = d(3 sec θ)






dx = 3 sec θ tan θdθ
√
p
x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3| tan θ| = 3 tan θ
Z
Z
3 sec θ
· 3 sec θ tan θdθ = 3 sec2 θdθ = 3 tan θ + C
=
3 tan θ
!
x
!! √
!
√
!
x2 − 9
x2 − 9
!!
!
tan θ =
!! θ
3
3
√
Z
x
x2 − 9 √ 2
√
= x −9+C
dx = 3
3
x2 − 9
Note that
Therefore
EXAMPLE 20: Find
Solution: We have
Z
x = 3 sec θ

Z
√
1
x2
−9
dx.
x = 3 sec θ



√
dx = 

x2 − 9

1
√
√
dx = d(3 sec θ)
dx = 3 sec θ tan θdθ
√
p
x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3| tan θ| = 3 tan θ
Z
Z
1
· 3 sec θ tan θdθ = sec θdθ = [Example 8] = ln | sec θ + tan θ| + C1
=
3 tan θ
√
x
x2 − 9
Note that sec θ = and tan θ =
, therefore
3
3
√
Z
2 − 9
√
x
1
x
+ C1 = ln |x + x2 − 9| + C
√
dx = ln | sec θ + tan θ| + C1 = ln +
3
3
x2 − 9
√
√
16






Section 6.2 Trigonometric Integrals and Substitutions
2010 Kiryl Tsishchanka
Appendix I
EXAMPLE: Find
Z
sin3 xdx.
Solution 1: We have
"
Z
Z
sin3 xdx = sin2 x · sin xdx =
sin2 x = u
sin xdx = dv
2 sin x cos xdx = du
− cos x = v
2
= sin x(− cos x) −
2
= − sin x cos x + 2
2
= − sin x cos x + 2
2
= − sin x cos x + 2
2
= − sin x cos x + 2
Z
Z
Z
Z
Z
(− cos x) · 2 sin x cos xdx
sin x cos2 xdx
sin x(1 − sin2 x)dx
(sin x − sin3 x)dx
sin xdx − 2
2
= − sin x cos x − 2 cos x − 2
therefore
3
so
Z
#
Z
Z
sin3 xdx
sin3 xdx
sin3 xdx = − sin2 x cos x − 2 cos x + C1
1
2
sin3 xdx = − sin2 x cos x − cos x + C
3
3
Z
Solution 2: We have
Z
3
sin xdx =
=−
Z
Z
2
sin x · sin xdx =
Z

cos x = u

 d cos x = du
(1 − cos x) sin xdx = 
 − sin xdx = du

2
1
1
(1 − u2 )du = −u + u3 + C = − cos x + cos3 x + C
3
3
17
sin xdx = −du






Section 6.2 Trigonometric Integrals and Substitutions
EXAMPLE: Find
Z
2010 Kiryl Tsishchanka
sinn xdx.
Solution: We have
"
Z
Z
sinn xdx = sinn−1 x · sin xdx =
sinn−1 x = u
sin xdx = dv
(n − 1) sinn−2 x cos xdx = du
− cos x = v
= sin
therefore
n
so
Z
n−1
x(− cos x) −
= − sin
n−1
= − sin
n−1
= − sin
n−1
= − sin
n−1
n
sin xdx = − sin
n−1
Z
#
(− cos x) · (n − 1) sinn−2 x cos xdx
x cos x + (n − 1)
x cos x + (n − 1)
x cos x + (n − 1)
Z
Z
Z
sinn−2 x cos2 xdx
sinn−2 x(1 − sin2 x)dx
(sinn−2 x − sinn x)dx
x cos x + (n − 1)
Z
sin
x cos x + (n − 1)
Z
sinn−2 xdx
n−2
xdx − (n − 1)
Z
sinn xdx
Z
n−1
1
n−1
sinn−2 xdx
x cos x +
sin xdx = − sin
n
n
In particular, if n = 4 we have
Z
Z
1 3
3
4
sin xdx = − sin x cos x +
sin2 xdx
4
4
Z
1 3
1
3
1
= − sin x cos x +
− sin x cos x +
dx
4
4
2
2
Z
n
3
3
1
= − sin3 x cos x − sin x cos x + x + C
4
8
8
Alternatively,
Z
Z 1 − cos 2x
2
2
Z
1
sin xdx = (sin x) dx =
(1 − cos 2x)2 dx
dx =
4
Z
Z 1
1
1
2
=
(1 − 2 cos 2x + cos 2x)dx =
1 − 2 cos 2x + (1 + cos 4x) dx
4
4
2
Z 3
1
3
1
1
1
− 2 cos 2x + cos 4x dx = x − sin 2x +
sin 4x + C
=
4
2
2
8
4
32
4
Z
2
2
18
Section 6.2 Trigonometric Integrals and Substitutions
2010 Kiryl Tsishchanka
Similarly, if n = 5 we have
Z
Z
1 4
4
5
sin xdx = − sin x cos x +
sin3 xdx
5
5
1 4
1
4
= − sin x cos x +
− sin2 x cos x +
5
5
3
1
4
1
− sin2 x cos x −
= − sin4 x cos x +
5
5
3
2
3
Z
sin xdx
2
cos +C1
3
1
4
8
= − sin4 x cos x −
sin2 x cos x −
cos x + C
5
15
15
Alternatively,
Z
Z
Z
5
4
sin xdx = sin x · sin xdx = (sin2 x)2 · sin xdx
=
Z

cos x = u

 d cos x = du
(1 − cos x) sin xdx = 
 − sin xdx = du

2
2
sin xdx = −du
=−
Z


Z

 = − (1 − u2 )2 du


2
1
2
1
(1 − 2u2 + u4 )du = −u + u3 − u5 + C = − cos x + cos3 x − cos5 x + C
3
5
3
5
19
Section 6.2 Trigonometric Integrals and Substitutions
2010 Kiryl Tsishchanka
Appendix II
x 1
1 − cos x cos
x
= ln 1. ln tan
−
= ln sin x sin x = ln |csc x − cot x|
2
sin x 
π π 
x π 1 − cos x +
x+
1
+
sin
x
1
sin
x
2  = ln 2 = ln = ln 2. ln tan
+
+
= ln tan 
cos x cos x cos x π
2 4
2
sin x +
2
= ln |sec x + tan x|
1
3.
ln
2
1 − cos x
1 + cos x
1
= ln
2
(1 − cos x)2
(1 − cos x)(1 + cos x)
1
(1 − cos x)2
(1 − cos x)2
= ln
1 − cos2 x
2
sin2 x
2
1 − cos x 1 − cos x 1
1 − cos x
1
= ln = ln
= · 2 ln sin x 2
sin x
2
sin x 1
cos x = ln |csc x − cot x|
−
= ln sin x sin x 1
4.
ln
2
1 + sin x
1 − sin x
1
= ln
2
(1 + sin x)2
(1 − sin x)(1 + sin x)
1
= ln
2
1 + sin x
cos x
2
1
= ln
2
1
= ln
2
(1 + sin x)2
1 − sin2 x
1
= ln
2
1 + sin x 1 + sin x 1
= ln = · 2 ln cos x 2
cos x 1
sin
x
= ln |sec x + tan x|
= ln +
cos x cos x 20
(1 + sin x)2
cos2 x