How do you find the antiderivative of int cos^3x dx∫cos3xdx?

1 Answer
Elijah T.
Feb 12, 2017

sinx - 1/3sin^3xsinx−13sin3x

Explanation:

Change cos^3xcos3x:

cos^3x = cos^2x*cosx = (1-sin^2)*cosxcos3x=cos2x⋅cosx=(1−sin2)⋅cosx

So we can write it as

int cosx*(1-sin^2x) dx∫cosx⋅(1−sin2x)dx

int cosx dx - int cosxsin^2x dx∫cosxdx−∫cosxsin2xdx

sinx - int cosxsin^2x dxsinx−∫cosxsin2xdx

Then we can solve using u substitution

u = sinxu=sinx and du = cosxdxdu=cosxdx

sinx - int u^2 dusinx−∫u2du (substituting u)

sinx - 1/3u^3sinx−13u3

sinx - 1/3sin^3xsinx−13sin3x (replug in x)

How to do these problems

I just learned how to solve these problems an hour ago at school! Basically, the gist is to use trig in order to "make" the derivative of one of the parts of the function. In this case, we only had cosine, so we had to make a sine component in order to use u substitution.

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