How do you integrate cos3(x3)dx?

1 Answer
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=3cos3(x3)13dx=3cos3(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)=1cos2(θ)=1sin2(θ).

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

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

3(1sin2(t))cos(t)dt=3(1s2)ds

Integrating term by term using snds=sn+1n+1 where n1.

3(1s2)ds=3(ss33)=3ss3

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

3ss3=3sin(x3)sin3(x3)+C