Download Trigonometric Integrals

Trigonometric Integrals
It is often necessary to evaluate integrals of the form
(sin x)m (cos x)n dx, where m and n are integers.
If one of the exponents, either m or n is and odd, there is a straightforward simplification.
Case 1: m = 2M + 1 is odd.
Then (sin x)m = (sin x)2M+1 = (sin2 x)M sin x = (1 − cos2 x)M sin x,
so our integral becomes
(sin x)m (cos x)n dx =
(1 − cos2 x)M sin x(cos x)n dx
and we may make the substitution u = cos x with du = − sin xdx to
get
1
(sin x)m (cos x)n dx =
(1 − cos2 x)M sin x(cos x)n dx = − (1 −
u2 )M un du
Example:
−
(sin x)11 (cos x)10 dx = − (1 − u2 )5 u10 du =
2
2 2
2 3
2 4
2 5
1 − 5u + 10(u ) − 10(u ) + 5(u ) − (u )
−
−
u10 du =
u10 − 5u12 + 10u14 − 10u16 + 5u18 − u20 du =
11
13
15
17
19
21
u
u
u
u
u
u
−5
+ 10
− 10
+5
−
11
13
15
17
19
21
+C =
cos11 x 5 cos13 x 2 cos15 x 10 cos17 x 5 cos 19x cos21 x
+
−
+
−
+
+C
−
11
13
5
17
19
21
2
Example:
−
(sin x)11 (cos x)−10 dx = − (1 − u2 )5 u−10 du =
2
2 2
2 3
2 4
2 5
1 − 5u + 10(u ) − 10(u ) + 5(u ) − (u )
−
u−10 du =
u−10 − 5u−8 + 10u−6 − 10u−4 + 5u−2 − u0 du =
−
−9
−7
−5
−3
−1
u
u
u
u
u
−5
+ 10
− 10
+5
−u +C =
−9
−7
−5
−3
−1
− −
C=
−9
7
−3
10 cos
cos x 5 cos x
+
− 2 cos−5 x +
9
7
3
x
− 5 cos−1 x − cos x +
5
10
1
sec9 x − sec7 x + 2sec5 x −
sec3 x + 5secx + cos x + C
9
7
3
3
Example: (sin x)−11(cos x)−10dx = (sin x)−12(cos x)−10 sin xdx =
− (1 − u2 )−6 u−10 du
can be done, but requires the method of Partial Fractions, which we
shall see later.
The situation is similar when the power of cos x is odd.
Example: (sin x)10 (cos x)11 dx = (sin x)10 (cos x)10 cos xdx =
(sin x)10 (cos2 x)5 cos xdx =
(sin x)10 (1 − sin2 x)5 cos xdx =
( letting u = sin x and du = cos xdx)
10
2 5
u (1−u ) du =
10
u
2
2 2
2 3
2 4
2 5
1 − 5u + 10(u ) − 10(u ) + 5(u ) − (u )
u10 − 5u12 + 10u14 − 10u16 + 5u18 − u20 du =
4
u11
u13
u15
u17
u19 u21
−5
+ 10
− 10
+5
−
+C =
11
13
15
17
19
21
5
2
10
5
1
1
sin11 x −
sin13 x + sin15 x −
sin17 x +
sin19 −
sin21 x + C
11
13
5
17
19
21
5
When both powers are odd, it is easiest to select the function with
the highest power for substitution:
Good Example:
(sin x)11 (cos x)3 dx =
(sin x)11 (cos x)2 cos xdx =
(sin x)11 (1 − sin2 x) cos xdx =
( letting u = sin x and du = cos xdx)
11
2
u (1 − u )du =
u
11
u11 u14
− u du =
−
+C =
11
14
13
1
1
sin11 x −
sin13 x + C
11
14
6
Bad Example:
(sin x)11 (cos x)3 dx =
(sin x)10 (cos x)3 sin xdx =
(1 − cos2 x)5 (cos x)3 sin xdx =
( letting u = cos x and du = − sin xdx)
(1 − u2 )5 u3 (−du) =
2
2 2
2 3
2 4
2 5
−1 + 5u − 10(u ) + 10(u ) − 5(u ) + (u ) u3 du =
3
5
7
9
−u + 5u − 10u + 10u − 5u
11
+u
13
du =
u6
u8
u10
u12 u14
u4
+5
− 10
+ 10
−5
+
+C =
−
4
6
8
10
12
14
5
5
5
1
1
cos12 x +
cos14 x + C
− cos4 x + cos6 x − cos8 x + cos10 x −
4
6
4
12
14
7
When neither power is odd, we need to use a double angle formula
from trigonometry:
cos 2x ≡ 2 cos2 x − 1 = 1 − 2 sin2 x
can be solved for cos2 x and sin2 x:
1 + cos 2x
1 − cos 2x
cos2 x =
and sin2 x =
2
2
Thus, if we want to integrate (sin x)m (cos x)n dx, where m and n
are even integers, we write m = 2M and n = 2N and we have:
(sin x)m (cos x)n dx =
(sin x)2M (cos x)2N dx =
(sin2 x) (cos2 x)N dx
1 − cos 2x M 1 + cos 2x N
dx =
2
2
M
N
−(M+N)
2
(1 − cos 2x) (1 + cos 2x) dx
8
M
1 + cos 2x
dx =
2
1
1
1
1 + cos 2xdx =
x + sin 2x + C =
2
2
2
Example:
cos2 xdx =
x 1
x + sin x cos x
x 1
+ sin 2x + C = + 2 sin x cos x + C =
+C
2 4
2 4
2
1 − cos 2x
dx =
2
1
1
1
1 − cos 2xdx =
x − sin 2x + C =
2
2
2
Example:
2
sin xdx =
x 1
x − sin x cos x
x 1
− sin 2x + C = − 2 sin x cos x + C =
+C
2 4
2 4
2
It should come as no surprise that
9
2
2
cos x + sin x dx = x + C
Example:
sin2 x cos2 xdx =
1 − cos 2x
2
1 + cos 2x
dx =
2
1
1
+
cos
4x
1
1
1 − cos2 2xdx =
1−
dx =
1 − cos 4xdx =
4
4
2
8
1
1
x sin 4x
x − sin 4x + C = −
+C
8
4
8
32
10
Products of powers of secant and tangent functions
secm x tann x can be expressed as a product of powers of sin x and
cos x, but it is often more convenient to use the identity tan2 x +
1 = sec2 x and the differentials d(tan x) = sec2 xdx and d(secx) =
secx tan xdx.
We assume that m and n are non-negative.
Case 1: The power m of secx is positive and even: then m = 2M
and we have:
secm x tann xdx =
(sec2 x)M tann xdx =
n
(sec2 x)M−1 t
xsec2 xdx =
(1 + tan2 x)M−1 tann xsec2 xdx = and we can make the substitution
u = tan x, du = sec2 xdx, and we get
M−1
M−1
2
n
2
2
1 + tan x
tan xsec xdx =
1+u
un du
11
Example:
10
5
sec x tan xdx =
1+u
1 + 4u2 + 6u4 + 4u6 + u8 u5 du =
2
4
u5 du =
u5 + 4u7 + 6u9 + 4u11 +
u13 du =
u8
u10
u12 u14
u6
+4
+6
+4
+
+C =
6
8
10
12
14
1 14
1 6 1 8 3 10 1 12
u + u + u + u +
u +C =
6
2
5
3
14
1
3
1
1
1
tan6 x + tan8 x + tan10 x + tan12 x +
tan14 x + C
6
2
5
3
14
12
If the power of secx is not even, but the power of n of tan x is odd,
so that n = 2N + 1, and m is positive, we can make the substitution
u = secx, du = secx tan xdx, and get
secm x tann xdx = secm−1 x tan2n xsecx tan xdx =
N
N
m−1
2
m−1
2
u
tan x du = u
sec x − 1 du =
u
m−1
N
2
u −1
Example:
du
5
5
sec x tan xdx =
4
2
2
u (u − 1) du =
1)du =
u9
u7 u5
8
6
4
−2
+
+C =
u − 2u + u du
9
7
5
2
1
1
sec9 x − sec7 x + sec5 x + C
9
7
5
13
u4 (u4 − 2u2 +
We are left with the cases where m is odd and n is even, all of which
can be reduced to the problem of finding the antiderivative of an odd
power of secx.
secx + tan x
dx =
Example: secxdx = secx
secx + tan x
(sec2 x + secx tan x)dx
sec2 x + secx tan x
dx =
=
secx + tan x
secx + tan x
d(secx + tan x)
= ln(secx + tan x) + C
secx + tan x
14
Example: I =
sec3 xdx
Using Integration by Parts, with
u = secx, dv = sec2 x, v = tan x, du = secx tan xdx, we get
I=
sec3 xdx =
udv = uv −
vdu =
secx tan x −
tan xsecx tan xdx = secx tan x −
tan2 xsecxdx =
secx tan x − (sec2 x − 1)secxdx =
secx tan x −
sec3 xdx +
secxdx = secx tan x − I +
secx tan x + ln |secx + tan x| + C so
2I = secx tan x + ln |secx + tan x| + C and
sec3 xdx =
secx tan x + ln |secx + tan x|
+C
2
15
secxdx =
Example:
tan2 xsecxdx appears in the previous calculation, and
is one of the simpler cases left:
I = secx tan x − tan2 xsecxdx gives us
tan2 xsecxdx = secx tan x − I =
secx tan x −
secx tan x + ln |secx + tan x|
+C =
2
secx tan x − ln |secx + tan x|
+C
2
16
Integrals of products of sine and cosine functions
with different arguments
The identities:
cos x cos y =
1
cos(x + y) + cos(x − y)
2
1
cos(x − y) − cos(x + y)
sin x sin y =
2
1
sin(x − y) + sin(x + y) may be used:
2
1
Example: cos 5t cos 7tdt =
(cos(5t + 7t) + cos(5t − 7t)) dt =
2
1 1
1
1
sin 12t + sin 2t + C =
(cos 12t + cos(−2t)) dt =
2
2 12
2
sin x cos y =
1
1
sin 12t + sin 2t + C
24
4
17