# Thread: Basic trigonometric integral question

1. ## 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.

2. ## Re: Basic trigonometric integral question

Looks good to me.

- Hollywood

3. ## Re: Basic trigonometric integral question

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

\displaystyle \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*}