This instruction material adopted of Calculus by Frank Ayres Jr
22
14. TRIGONOMETRIC INTEGRALS
THE FOLLOWING IDENTITIES are employed to find the trigonometric integrals
of this chapter. 1.
1 x
cos x
sin
2 2
2.
x sec
x tan
1
2 2
3.
x csc
x cot
1
2 2
4.
x 2
cos 1
x sin
2 1
2
5.
x 2
cos 1
x cos
2 1
2
6.
x 2
sin x
cos x
sin
2 1
7.
y x
sin y
x sin
y cos
x sin
2 1
8.
y x
cos y
x cos
y sin
x sin
2 1
9.
y x
cos y
x cos
y cos
x cos
2 1
10.
x sin
2 x
cos 1
2 1
2
11.
x cos
2 x
cos 1
2 1
2
12.
x cos
1 x
sin 1
2 1
SOLVED PROBLEMS SINES AND COSINES
1. C
x x
dx x
dx x
2 sin
2 cos
1 sin
4 1
2 1
2 1
2
2. C
x x
dx x
dx x
6 sin
6 cos
1 3
cos
12 1
2 1
2 1
2
3. C
x x
dx x
x dx
x x
dx x
3 8
1 2
2 3
cos cos
sin cos
1 sin
sin sin
4. dx
x x
dx x
x dx
x cos
sin 1
cos cos
cos
2 2
4 3
dx x
x dx
x x
dx x
cos sin
cos sin
2 cos
4 2
C x
x x
3 5
1 3
3 2
sin sin
sin
5. dx
x x
x dx
x x
x dx
x x
cos sin
1 sin
cos cos
sin cos
sin
2 2
2 2
3 2
C x
x dx
x x
dx x
x
5 5
1 3
3 1
4 2
sin sin
cos sin
cos sin
This instruction material adopted of Calculus by Frank Ayres Jr
23
15. TRIGONOMETRIC SUBSTITUTIONS
AN INTEGRAND, which contains one of the forms
2 2
2
u b
a ,
2 2
2
u b
a , or
2 2
2
a u
b but no other irrational factor, may be transformed into another involving
trigonometric functions of a new variable as follows: For
Use To obtain
2 2
2
u b
a z
b a
u sin
z a
z a
cos sin
1
2 2
2 2
u b
a
z b
a u
tan
z a
z a
sec tan
1
2 2
2 2
a u
b
z b
a u
sec
z a
z a
tan 1
sec
2
In each case, integration yields an expression in the variable z. The corresponding expression in the original variable may be obtained by the use of a right triangle as
shown in the solved problems below.
SOLVED PROBLEMS
1. Find
2 2
4 x
x dx
Let x = 2 tan z; then dx = 2 sec
2
z dz and z
x sec
2 4
2
dz z
z z
z dz
z x
x dx
2 2
2 2
2
tan sec
4 1
sec 2
tan 4
sec 2
4 C
x x
C z
dz z
z 4
4 sin
4 1
cos sin
4 1
2 2
2. Find
dx x
x 4
2 2
Let x = 2 sec z; then dx = 2 sec z tan z dz and z
x tan
2 4
2
dz z
dz z
z z
z dx
x x
3 2
2 2
sec 4
tan sec
2 tan
2 sec
4 4
C z
z z
z tan
sec ln
2 tan
sec 2
C x
x x
x 4
ln 2
4
2 2
2 1
This instruction material adopted of Calculus by Frank Ayres Jr
24
3. Find
dx x
x
2
4 9
Let
z x
sin
2 3
; then
dz z
dx cos
2 3
and z
x cos
3 4
9
2
dz z
z dz
z z
z dx
x x
sin cos
3 cos
sin cos
3 4
9
2 2
3 2
3 2
dz z
dz z
dz z
z sin
3 csc
3 sin
sin 1
3
2
C z
z z
cos 3
cot csc
ln 3
C x
x x
2 2
4 9
4 9
3 ln
3
16. INTEGRATION BY PARTIAL FRACTIONS