1 answer

How do you find the integral of #int [sin(3x)]^4 dx#?

Question:

How do you find the integral of #int [sin(3x)]^4 dx#?

Answers

Use power reduction formulas twice.

Explanation:

#cos(2x) = cos^2x-sin^2x#
#cos(2x) = 1-2sin^2x#
#cos(2x) = 2cos^2x-1#

So,
#sin^2x = 1/2(1-cos(2x))#
and,
#cos^2x = 1/2(1+cos(2x))#

#[sin(3x)]^4 = [color(blue)(sin^2 3x)]^2#

# = [color(blue)(1/2(1-cos6x))]^2#

# = 1/4[1-2cos6x+color(red)(cos^2 6x)]#

# = 1/4[1-2cos6x+(color(red)(1/2(1+cos12x)))]#

# = 1/8[2(1-2cos6x)+(1+cos12x))]#

# = 1/8[3-4cos6x+cos12x]#

The last expression can be integrated term by term using substitution for the two terms involving cosine.

.

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