# Basic trigonometric integral question

• Feb 22nd 2013, 05:06 AM
DonGorgon
Basic trigonometric integral question
Just wondering about the validity of this answer obtained via using an identity?
A second opinion would be much appreciated! Thanks.
• Feb 22nd 2013, 06:23 AM
hollywood
Re: Basic trigonometric integral question
Looks good to me.

- Hollywood
• Feb 22nd 2013, 06:58 AM
Prove It
Re: Basic trigonometric integral question
Quote:

Originally Posted by DonGorgon
Just wondering about the validity of this answer obtained via using an identity?
A second opinion would be much appreciated! Thanks.

Even though this solution looks fine, it is NOT the most efficient method in this case. I always tell my students to look for obvious substitutions.

\displaystyle \begin{align*} \int{\sin^3{(x)}\,dx} &= -\int{-\sin{(x)}\sin^2{(x)}\,dx} \\ &= -\int{-\sin{(x)}\left[ 1 - \cos^2{(x)} \right] dx} \end{align*}

Now let \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*} and the integral becomes

\displaystyle \begin{align*} -\int{-\sin{(x)}\left[ 1 - \cos^2{(x)} \right] dx} &= -\int{ 1 - u^2 \, du } \\ &= - \left( u - \frac{u^3}{3} \right) +C \\ &= -\cos{(x)} + \frac{\cos^3{(x)}}{3} + C \end{align*}