Powers of Trig Functions
Powers of Sin and Cos
Example:
Evaluate
sin5x dx
we write
sin5x dx
=
sin4x
sinx dx
=
(sin2
x)2sinx dx
=
(1-cos2x)2
sinx dx u
= cosx du = -sinx dx
= - (1
- u2)2 du
= - [1
- 2u2 + u4] du
= -u + 2/3 u3 -
1/5 u5 + C
= -cosx + 2/3 cos3x
- 1/5 cos5x + C
Exercise
Evaluate
cos7x dx
Example
Evaluate
sin4x dx
Solution
We write
sin4x dx
= (sin2x)2 dx
= (1/2 - 1/2 cos(2x))2 dx
=
[1/4 - 1/2 cos(2x) + 1/4 cos2(2x)] dx
For the second
integral let
u = 2x
dx = 1/2du
The integral becomes
1/4 x - 1/4 sin(2x) + 1/4
[1/2 + 1/2 cos(4x)] dx
Let u = 4x
dx = 1/4du
= 1/4 x - 1/4 sin(2x) + 1/8 x + 1/32 sin(4x) + C
We see that if the power is odd we can pull out one of the sin functions
and convert the other to an expression involving the cos function only.
Then use
u = cos x.
If the power is even, we must use the trig identities
sin2x = 1/2 - 1/2 cos(2x)
and
cos2x = 1/2 + 1/2 cos(2x) |
This method will always work and is always long and tedious.
Mixed Powers of Sin and Cos
Example
Evaluate
sin2x cos3x dx
We pull out a cos x and convert
the cos2 x to 1 - sin2 x:
sin2x (1 - sin2 x) cosx dx
Let u = sin x,
du = cos x dx
We have
u2(1 - u2)du
=
u2 - u4
du = 1/3 u3 - 1/5 u5 + C
= 1/3 sin3x -
1/5 sin5x + C
Exercise
sin5x cos2x dx
Powers of Tangents and Secants
To integrate powers of tangents and secants we use the formula
tan2 x
= sec2 x - 1
Example
Evaluate
tan4 x dx
Solution
We write
tan2 x tan2 x dx
=
tan2 x (sec2 x - 1)
= tan2 x sec2 x dx
- tan2 x dx
= tan2 x sec2 x dx
- (sec2
x - 1) dx
For the first integral let
u = tan x du = sec2 x dx.
We have
u2 du - tanx
+ x
= 1/3 u3 - tanx
+ x + C
= 1/3 tan3 x - tan x
+ x + C
Exercise
sec5
x tan3 x dx.
Mixed Angles
We have the following formulas:
sin(mx) sin(nx) = 1/2[cos[(m - n)x] - cos[(m + n)x]]
sin(mx) cos(nx) = 1/2[sin[(m - n)x] + sin[(m + n)x]]
cos(mx) cos(nx) = 1/2[cos[(m - n)x] + cos[(m + n)x]]
|
Example
sin(3x)sin(4x)dx
= 1/2
[cos(-x) - cos(7x)] dx
We integrate the first by letting u = -x and the second by letting
u = 7x to get
1/2 -sin(-x) +
1/7 sin(7x) + C
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