Originally Posted by

**Prove It** If your integral is $\displaystyle \begin{align*} \int{\cot^5{(x)}\sin^4{(x)}\,dx} \end{align*}$, I would do...

$\displaystyle \begin{align*} \int{\cot^5{(x)}\sin^4{(x)}\,dx} &= \int{ \frac{\cos^5{(x)}}{\sin^5{(x)}} \cdot \sin^4{(x)} \,dx } \\ &= \int{ \frac{\cos^5{(x)}}{\sin{(x)}} \,dx } \\ &= \int{ \frac{\cos{(x)} \left[ \cos^2{(x)} \right] ^2 }{ \sin{(x)} } \, dx } \\ &= \int{ \frac{\left[ 1 - \sin^2{(x)} \right] ^2 }{ \sin{(x)} } \cdot \cos{(x)} \, dx } \end{align*}$

Now with the substitution $\displaystyle \begin{align*} u = \sin{(x)} \implies du = \cos{(x)} \, dx \end{align*}$ we have

$\displaystyle \begin{align*} \int{ \frac{\left[ 1 - \sin^2{(x)} \right] ^2}{\sin{(x)}} \cdot \cos{(x)}\,dx } &= \int{ \frac{\left( 1 - u^2 \right) ^2}{u} \, du } \\ &= \int{ \frac{1 - 2u^2 + u^4}{u} \, du } \\ &= \int{\frac{1}{u} - 2u + u^3 \, du} \\ &= \ln{|u|} - u^2 + \frac{1}{4}u^4 + C \\ &= \ln{ \left| \sin{(x)} \right| } - \sin^2{(x)} + \frac{1}{4}\sin^4{(x)} + C \end{align*}$