# integral cos (x^3) dx is?

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Evaluate `int cos^3x dx` :

Rewrite as `intcos^3xdx=int(cos^2x)cosxdx`

`=int(1-sin^2x)cosxdx`

Let `u=sinx,du=cosxdx` Then

`int(1-sin^2x)cosxdx=int(1-u^2)du`

`=intdu-intu^2du`

`=u-1/3u^3+C_1` Substituting for `u` we get:

`=sinx-1/3sin^3x+C`

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`intcos^3xdx=sinx-1/3sin^3x+C`

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