Evaluate #I=\int\cos^4(x)dx#?

Symbolab says to use integration by parts.
However, this question originates in a "Trigonometric Integrals" packet... so how would I go about solving this?

1 Answer
Øko
Apr 9, 2018

#I=1/32(12x+sin(4x)+8sin(2x))+C#

Explanation:

We want to solve

#I=intcos^4(x)dx#

I always looking to reduce the powers of the integrand,
for integrals of this type

You could use the identity

#color(blue)(cos^2(x)=1/2(1+cos(2x))#

Thus

#I=int(cos^2(x))^2dx#

#color(white)(I)=1/4int(1+cos(2x))^2dx#

#color(white)(I)=1/4int1+cos^2(2x)+2cos(2x)dx#

#color(white)(I)=1/4int1+1/2(1+cos(4x))+2cos(2x)dx#

#color(white)(I)=1/4int3/2+1/2cos(4x)+2cos(2x)dx#

Which is much nicer to integrate

#I=1/4(3/2x+1/8sin(4x)+sin(2x))+C#

#color(white)(I)=1/32(12x+sin(4x)+8sin(2x))+C#

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