# Math Help - messy integral (?)

1. ## messy integral (?)

could somebody help me integrate the following please:

(sin x)^4 (cos x)^5 dx

(the limits are pi/2 and 0 but i can solve that once the initial integration has been done)

im not sure how to split it up to make it easier, because when i do use trig identities it becomes very messy....

cheers

2. $\int_0^{\pi /2} {\sin ^4 x\cos ^5 x\,dx} = \int_0^{\pi /2} {\sin ^4 x\left( {\cos ^2 x} \right)^2 \cos x\,dx} .$

Since you know that $\cos^2x=1-\sin^2x,$ you just replace it and make the obvious substitution. (Don't forget to get the new integration limits for your new variable.)

3. Originally Posted by nmanik90
could somebody help me integrate the following please:

(sin x)^4 (cos x)^5 dx

(the limits are pi/2 and 0 but i can solve that once the initial integration has been done)

im not sure how to split it up to make it easier, because when i do use trig identities it becomes very messy....

cheers
Put $\cos^5 x = \cos^4 x \cos x = (1 - \sin^2 x)^2 \cos x$. Now make the substitution $u = \cos x$.

4. There is a general way for $\int_0^{\pi/2} \sin^n x \cos^m x ~ dx$. Here