1.6. TRIGONOMETRIC INTEGRALS AND TRIG.

SUBSTITUTIONS 26

1.6. Trigonometric Integrals and Trigonometric

Substitutions

1.6.1. Trigonometric Integrals. Here we discuss integrals of pow-

ers of trigonometric functions. To that end the following half-angle
identities will be useful:
1
sin2 x = (1 − cos 2x) ,
2
1
cos2 x = (1 + cos 2x) .
2

Remember also the identities:

sin2 x + cos2 x = 1 ,
sec2 x = 1 + tan2 x .

1.6.1.1. Integrals of Products of Sines and Cosines. We will study

now integrals of the form
Z
sinm x cosn x dx ,

including cases in which m = 0 or n = 0, i.e.:

Z Z
n
cos x dx ; sinm x dx .

The simplest case is when either n = 1 or m = 1, in which case the

substitution u = sin x or u = cos x respectively will work.
Z
Example: sin4 x cos x dx = · · ·

(u = sin x, du = cos x dx)

Z
4 u5 sin5 x
··· = u du = +C = +C .
5 5

More generally if at least one exponent is odd then we can use the
identity sin2 x+cos2 x = 1 to transform the integrand into an expression
containing only one sine or one cosine.
 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 27

Example:
Z Z
sin x cos x dx = sin2 x cos2 x cos x dx
2 3

Z
= sin2 x (1 − sin2 x) cos x dx = · · ·

(u = sin x, du = cos x dx)

Z Z
· · · = u (1 − u ) du = (u2 − u4 ) du
2 2

u 3 u5
= − +C
3 5
sin3 x sin5 x
= − +C .
3 5

If all the exponents are even then we use the half-angle identities.
Example:
Z Z
2 2 1
sin x cos x dx = 2
(1 − cos 2x) 12 (1 + cos 2x) dx
Z
1
= (1 − cos2 2x) dx
4
Z
1
= (1 − 12 (1 + cos 4x)) dx
4
Z
1
= (1 − cos 4x) dx
8
x sin 4x
= − +C.
8 32

1.6.1.2. Integrals of Secants and Tangents. The integral of tan x

can be computed in the following way:
Z Z Z
sin x du
tan x dx = dx = − = − ln |u| + C = − ln | cos x| + C ,
cos x u
where u = cos x. Analogously
Z Z Z
cos x du
cot x dx = dx = = ln |u| + C = ln | sin x| + C ,
sin x u
where u = sin x.
 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 28

The integral of sec x is a little tricky:

Z Z Z
sec x (tan x + sec x) sec x tan x + sec2 x
sec x dx = dx = dx =
sec x + tan x sec x + tan x
Z
du
= ln |u| + C = ln | sec x + tan x| + C ,
u
where u = sec x + tan x, du = (sec x tan x + sec2 x) dx.
Analogously:
Z
csc x dx = − ln | csc x + cot x| + C .

More generally an integral of the form

Z
tanm x secn x dx

can be computed in the following way:

(1) If m is odd, use u = sec x, du = sec x tan x dx.

(2) If n is even, use u = tan x, du = sec2 x dx.
Z
Example: tan3 x sec2 x dx = · · ·

Since in this case m is odd and n is even it does not matter which
method we use, so let’s use the first one:

(u = sec x, du = sec x tan x dx)

Z Z
··· = tan x sec x tan x sec x dx = (u2 − 1)u du
2

| {z } | {z } | {z }
u2 −1 u du
Z
= (u3 − u) du

u4 u2
= − +C
4 2
= 1
4
sec4 x − 12 sec2 x + C .

Next let’s solve the same problem using the second method:
 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 29

(u = tan x, du = sec2 x dx)

Z Z
u4
tan x sec x dx = u3 du =
3 2
+C = 1
4
tan4 x + C .
4
| {z } | {z }
u3 du

Although this answer looks different from the one obtained using the
first method it is in fact equivalent to it because they differ in a con-
stant:
1
4
tan4 x = 14 (sec2 x − 1)2 = 14 sec4 x − 12 sec2 x + 14 .
| {z }
previous answer

1.6.2. Trigonometric Substitutions. Here we study substitu-

tions of the form x = some trigonometric function.
Z √
Example: Find 1 − x2 dx.

Answer : We make x = sin t, dx = cos t dt, hence

√ p √
1 − x2 = 1 − sin2 t = cos2 t = cos t ,
and
Z √ Z
1− x2 dx = cos t cos t dt
Z
= cos2 t dt
Z
1
= 2
(1 + cos 2t) dt (half-angle identity)
t sin 2t
= + +C
2 4
t 2 sin t cos t
= + +C (double-angle identity)
2 4
p
t sin t 1 − sin2 t
= + +C
2 2
√
sin−1 x x 1 − x2
= + +C .
2 2

The following substitutions are useful in integrals containing the

following expressions:
 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 30

expression substitution identity

a2 − u2 u = a sin t 1 − sin2 t = cos2 t
a2 + u2 u = a tan t 1 + tan2 t = sec2 t
u2 − a2 u = a sec t sec2 t − 1 = tan2 t

So for instance, if an integral contains the expression a2 −u2 , we may

try the substitution u = a sin t and use the identity 1 − sin2 t = cos2 t
in order to transform the original expression in this way:

a2 − u2 = a2 (1 − sin2 t) = a2 cos2 t .

Example:

Z Z
x3 sin3 t cos t
√ dx = 27 p dt (x = 3 sin t)
9 − x2 1 − sin2 t
Z
= 27 sin3 t dx
Z
= 27 (1 − cos2 t) sin t dx
µ ¶
cos3 t
= 27 − cos t + +C
3
³ p ´
= 27 − 1 − sin2 t + 13 (1 − sin2 t)3/2 + C
√
= −9 9 − x2 + 13 (9 − x2 )3/2 + C .

where x = 3 sin t, dx = 3 cos t dt.

 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 31

Example:
Z √ Z q
9
9 + 4x dx = 2
2
4
+ x2 dx (x = 32 tan t)
Z p
3 3
=2 1 + tan2 t sec2 t dt
2 2
Z
9
= sec3 t dt
2
9
= (sec t tan t + ln | sec t + tan t|) + C1
4µ ¯ ¯¶
q ¯ q ¯
9 2 4 2 ¯ 2 x + 1 + 4 x2 ¯ + C1
= x 1 + x + ln ¯3 ¯
4 3 9 9

√
x 9 + 4x2 9 √
= + ln |2x + 9 + 4x2 | + C .
2 4
where x = 32 tan t, dx = 32 sec2 t dt
Example:
Z √ 2 Z √ 2
x −1 sec t − 1
dx = sec t tan t dt (x = sec t)
x sec t
Z
= tan2 t dt
= tan t − t + C
√
= sec2 t − 1 − t + C
√
= x2 − 1 − sec−1 x + C .
where x = sec t, dx = sec t tan t dt.