Objective:
To compute the integral of a function involving the power of sine or the power of cosine
Method
In order to solve the integral of the power of sine or the power of cosine we have to reduce their power. We use the following formulas:
sin²x = 1/2 (1-cos2x )
cos²x = 1/2 ( 1 + cos2x )
Example I
Solve ∫ sin²xdx
Let's substitute sin²x by 1/2 ( 1-cos2x )
∫sin²xdx = ∫ [1/2 ( 1-cos2x )] dx
= 1/2 ∫ ( 1-cos2x ) dx
= 1/2 [ ∫ dx - ∫ cos2x dx ]
Let's solve ∫ cos2x dx
Let's use the substitution method by writing u = 2x
Then du = 2xdx dx = du/2
∫ cos2x dx = ∫ cosu.du/2
= 1/2 ∫ cosudu
= 1/2 sinu + C
= 1/2 sin2x + C
Let's solve the expression between brackets:
∫sin²xdx = 1/2 ( x - 1/2 sin2x ) + C
= 1/2 x - 1/4 sin2x + C
Example II Evaluate ∫ cos⁴x
The technique used to solve the integral of the power of cosine and the power of sine consists in reducing the power of these functions.
Let's first reduce the power of cos⁴x
cos⁴x = ( cos²x )²
Let's reduce the power of cos²x by using the formula cos²x = 1/2 ( 1 + cos2x )
cos⁴x = [ 1/2 ( 1 + cos2x ) ]²
= 1/4 ( 1 + 2cos2x + cos²2x )
= 1/4 + 1/2cos2x + 1/4cos²2x
Let's reduce the power of cos²2x:
cos²x = 1/2 ( 1 + cos2x ). By analogy cos²2x = 1/2 ( 1 + cos4x ). Then 1/4cos²2x = 1/8 + 1/8cos4x
Therefore cos⁴x = 1/4 + 1/2cos2x + 1/8 + 1/8cos4x
cos⁴x = 3/8 + 1/2cos2x + 1/8cos4x
∫cos⁴xdx = ∫( 3/8 + 1/2cos2x + 1/8cos4x )dx
= 3/8∫dx + 1/2∫ cos2xdx + 1/8∫ cos4xdx
In the example above we found ∫cos2xdx = 1/2 sin2x + C. By analogy ∫cos4xdx = 1/4 sin4x + C
∫cos⁴xdx = 3/8x + 1/2 ( 1/2 sin2x ) + 1/8 ( 1/4 sin4x ) + C
= 3/8x + 1/4 sin2x + 1/32 sin4x + C
Practice
Evaluate:
1) ∫cos²xdx
2) ∫sin⁴xdx
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