# definite integral!

• Apr 9th 2012, 02:40 AM
lawochekel
definite integral!
evaluate $\displaystyle \int\sin^2x\cos^6xdx$ having a limit $\displaystyle x=\frac{\pi}{2}, x=0$

pls help on the problem above, thanks.
• Apr 9th 2012, 02:45 AM
Prove It
Re: definite integral!
Quote:

Originally Posted by lawochekel
evaluate $\displaystyle \int\sin^2x\cos^6xdx$ having a limit $\displaystyle x=\frac{\pi}{2}, x=0$

pls help on the problem above, thanks.

Maybe try Integration by Parts with \displaystyle \displaystyle \begin{align*} u = \sin{x} \end{align*} and \displaystyle \displaystyle \begin{align*} dv = \sin{x}\cos^6{x}\,dx \end{align*}...
• Apr 9th 2012, 02:54 AM
princeps
Re: definite integral!
$\displaystyle \text{ Since }: \sin^2 x=1-\cos^2 x \text{ it follows :}$

$\displaystyle I= \int\limits_{0}^{\pi /2} \cos^6 x\,dx -\int\limits_{0}^{\pi /2} \cos^8 x\,dx$

Now , use reduction formula for cosine several times .
• Apr 9th 2012, 03:28 AM
Prove It
Re: definite integral!
Quote:

Originally Posted by princeps
$\displaystyle \text{ Since }: \sin^2 x=1-\cos^2 x \text{ it follows :}$

$\displaystyle I= \int\limits_{0}^{\pi /2} \cos^6 x\,dx -\int\limits_{0}^{\pi /2} \cos^8 x\,dx$

Now , use reduction formula for cosine several times .

My way should still work...

\displaystyle \displaystyle \begin{align*} \int{\sin^2{x}\cos^6{x}\,dx} &= -\frac{\sin{x}\cos^7{x}}{7} - \int{-\frac{\cos^8{x}}{7}\,dx} \\ &= -\frac{\sin{x}\cos^7{x}}{7} + \int{\frac{\cos^8{x}}{7}\,dx} \end{align*}

and now use the double angle formula to simplify \displaystyle \displaystyle \begin{align*} \cos^8{x} \end{align*}.