trigonometric integral: transformation required?

Hi. It's getting quite late at night, but I still have to figure out this integral:

$\displaystyle \int{cos^{3}x\,sin^{4}x\,dx$

I believe this is supposed to be solved with integration by parts, but I think there needs to be some kind of regrouping or transformation to make it work. I'm not seeing it. Could somebody help me out?

Re: trigonometric integral: transformation required?

Quote:

Originally Posted by

**infraRed** Hi. It's getting quite late at night, but I still have to figure out this integral:

$\displaystyle \int{cos^{3}x\,sin^{4}x\,dx$

I believe this is supposed to be solved with integration by parts, but I think there needs to be some kind of regrouping or transformation to make it work. I'm not seeing it. Could somebody help me out?

Never go for integration by parts when there is a simple substitution.

$\displaystyle \displaystyle \begin{align*} \int{\cos^3{(x)}\sin^4{(x)}\,dx} &= \int{\cos{(x)}\cos^2{(x)}\sin^4{(x)}\,dx} \\ &= \int{\cos{(x)}\left[ 1 - \sin^2{(x)} \right] \sin^4{(x)}\,dx} \\ &= \int{\cos{(x)}\left[ \sin^4{(x)} - \sin^6{(x)} \right] dx } \end{align*}$

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

$\displaystyle \displaystyle \begin{align*} \int{\cos{(x)}\left[ \sin^4{(x)} - \sin^6{(x)} \right] dx} &= \int{u^4 - u^6\,du} \\ &= \frac{u^5}{5} - \frac{u^7}{7} + C \\ &= \frac{\sin^5{(x)}}{5} - \frac{\sin^7{(x)}}{7} + C \end{align*}$