How do you integrate ∫cos3(x3)dx?

1 Answer
mason m
Jan 9, 2017

3sin(x3)−sin3(x3)+C

Explanation:

First let t=x3. This implies that dt=13dx. We then see that:

∫cos3(x3)dx=3∫cos3(x3)13dx=3∫cos3(t)dt

To do this, split up cos3(t) into cos2(t)cos(t) and then rewrite cos2(t) using the identity sin2(θ)+cos2(θ)=1⇒cos2(θ)=1−sin2(θ).

3∫cos3(t)dt=3∫cos2(t)cos(t)dt=3∫(1−sin2(t))cos(t)dt

Now let s=sin(t), so ds=cos(t)dt. Luckily we already have this in the integrand!

3∫(1−sin2(t))cos(t)dt=3∫(1−s2)ds

Integrating term by term using ∫snds=sn+1n+1 where n≠−1.

3∫(1−s2)ds=3(s−s33)=3s−s3

From s=sin(t) and t=x3 we see that s=sin(x3). Also add the constant of integration.

3s−s3=3sin(x3)−sin3(x3)+C

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