Section 7.2 Trigonometric Integrals
Ruipeng Shen
November 18
1
Powers of Sine and Cosine
Z
Example 1. Evaluate
Solution
cos3 x dx.
We can rewrite the integral into
Z
Z
Z
3
2
cos x dx = cos x · cos x dx = (1 − sin2 x) cos x dx
Apply the u-substitution u = sin x =⇒ du = cos x dx, we have
Z
Z
sin3 x
u3
+ C = sin x −
+ C.
cos3 x dx = (1 − u2 ) du = u −
3
3
Z
Example 2. Evaluate
sin5 x cos2 x dx.
Solution Using the same idea as the previous example, we have
Z
Z
sin5 x cos2 x dx = (sin2 x)2 cos2 x sin x dx
Z
= (1 − cos2 x)2 cos2 x sin x dx
let u = cos x =⇒ du = − sin x dx
Z
= (1 − u2 )2 u2 · (−du)
Z
=
−u2 + 2u4 − u6 du
u5
u7
u3
+2·
−
+C
3
5
7
1
2
1
= − cos3 x + cos5 x − cos7 x + C.
3
5
7
Z
Example 3. Evaluate
cos2 x sin2 x dx.
=−
1
Solution
By the half-angle identities
cos u sin u =
sin 2u
,
2
sin2 u =
1 − cos 2u
,
2
we have
cos2 x sin2 x = (cos x sin x)2 =
cos2 u =
1 + cos 2u
;
2
(1)
1 − cos 4x
sin2 2x
=
.
4
8
Thus we can integrate
Z
Z
1 − cos 4x
2
2
cos x sin x dx =
dx
8
Z
Z
1
1
dx −
cos 4x dx
=
8
8
Z
x 1
du
= −
cos u ·
8 8
4
x
1
= −
sin u + C
8 32
x sin 4x
+ C.
= −
8
32
Z
Strategy for Evaluating
sinm x cosn dx
let u = 4x =⇒ du = 4 dx
• If the power of cosine is odd (n = 2k + 1), save one cosine factor and use cos2 x = 1 − sin2 x
to express the remaining factors in term of sine:
Z
Z
Z
sinm x cosn xdx = sinm x(cos2 x)k cos x dx = sinm x(1 − sin2 x)k cos x dx,
then substitute u = sin x.
• If the power of sine is odd (m = 2k + 1), save one sine factor and use sin2 x = 1 − cos2 x
to express the remaining factors in term of cosine:
Z
Z
Z
m
2
n
k
n
sin x cos xdx = (sin x) cos x sin x dx = (1 − cos2 x)k cosn x sin x dx,
then substitute u = cos x.
• If the powers of both sine and cosine are even, use the half-angle identities (1).
2
Powers of Tangent and Secant
Z
Example 4. Evaluate
tan6 x sec4 x dx.
2
Solution
By the identity
sec2 x = 1 + tan2 x
and the substitution u = tan x =⇒ du = sec2 x dx, we have
Z
Z
tan6 x sec4 x dx = tan6 x(tan2 x + 1) sec2 x dx
Z
= u6 (1 + u2 ) du
u7
u9
+
+C
7
9
1
1
= tan7 x + tan9 x + C.
7
9
=
Z
Example 5. Calculate
Solution
tan5 θ sec7 θ dθ.
By the identity
sec2 θ = 1 + tan2 θ
and the substitution u = sec θ =⇒ du = sec θ tan θ dθ, we have
Z
Z
tan5 θ sec7 θ dθ = (tan2 θ)2 sec6 θ(sec θ tan θ)dθ
Z
= (sec2 θ − 1)2 sec6 θ(sec θ tan θ)dθ
Z
= (u2 − 1)2 u6 du
Z
=
u10 − 2u8 + u6 du
u9
u7
u11
−2·
+
+C
11
9
7
1
2
1
= sec11 θ − sec9 θ + sec7 θ + C.
11
9
7
=
Z
Strategy for Evaluating
tanm x secn x dx
• If the power of secant is even (n = 2k), save a factor of sec2 x and use sec2 x = 1 + tan2 x
to express the remaining factors in term of tan x:
Z
Z
Z
tanm x secn xdx = (tanm x)(sec2 x)k−1 sec2 x dx = (tanm x)(1+tan2 x)k−1 sec2 x dx,
then substitute u = tan x.
• If the power of tangent is odd (m = 2k + 1), save one factor of sec x tan x and use tan2 x =
sec2 x − 1 to express the remaining factors in term of sec x:
Z
Z
m
n
tan x sec xdx = (tan2 x)k secn−1 x sec x tan x dx
Z
= (sec2 x − 1)k secn−1 x(sec x tan x) dx,
then substitute u = sec x.
3
More Basic formulas on Trigonometric Integrals
Z
Z
tan x dx = ln | sec x| + C
sec x dx = ln | sec x + tan x| + C
Z
Z
cot x dx = ln | sin x| + C
csc x dx = ln | csc x − cot x| + C
Z
Example 6. Find
Solution
tan3 x dx
By the identity tan2 x = sec2 x − 1 and basic formula above, we have
Z
Z
tan3 x dx = (sec2 x − 1) tan x dx
Z
Z
2
= tan x sec x dx − tan x dx
let u = tan x du = sec2 x dx
Z
= u du − ln | sec x|
u2
− ln | sec x| + C
2
tan2 x
− ln | sec x| + C
=
2
=
Z
Example 7. Find
Solution
sec3 x dx
By integration by parts, we have
Z
Z
3
sec x dx = sec x(tan x)0 dx
Z
= sec x tan x − (tan x)(sec x)0 dx
Z
= sec x tan x − sec x tan2 x dx
Z
= sec x tan x − sec x(sec2 x − 1) dx
Z
Z
3
= sec x tan x − sec x dx + sec x dx
Thus we have an equation
Z
Z
2 sec3 x dx = sec x tan x + sec x dx = sec x tan x + ln | sec x + tan x| + C
Z
1
1
=⇒ sec3 x dx = sec x tan x + ln | sec x + tan x| + C
2
2
4
More
R Formulas (Optional) To evaluate the integrals of
or sin mx sin nx dx, use the corresponding identity
R
sin mx cos nx dx,
1
sin A cos B = [sin(A − B) + sin(A + B)]
2
1
sin A sin B = [cos(A − B) − cos(A + B)]
2
1
cos A cos B = [cos(A − B) + cos(A + B)]
2
Z
Example 8. Evaluate
Solution
sin 4x cos 5x dx.
By the identity above,
Z
Z
1
sin 4x cos 5x dx =
[sin(−x) + sin(9x)] dx
2
1
1
cos x − cos 9x + C
=
2
9
1
1
= cos x −
cos 9x + C.
2
18
5
R
cos mx cos nx dx