# Integral

• Feb 6th 2008, 03:52 AM
slevvio
Integral
Hello, some of you may remember me from the previous question.

$\displaystyle \int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx$

Hence evaluate

$\displaystyle \int^{\frac{\pi}{2}}_{0} xsin^2xcos^2xdx$

After messing about with it I can never seem to get it into a nicer form... thanks any help would be appreciated (Yes)
• Feb 6th 2008, 04:55 AM
galactus
You could try rewriting your integral.

$\displaystyle xsin^{2}xcos^{2}x=\frac{x^{2}}{8}-\frac{x^{2}cos(4x)}{8}$
• Feb 6th 2008, 05:24 AM
colby2152
Quote:

Originally Posted by slevvio
Hello, some of you may remember me from the previous question.

$\displaystyle \int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx$

Hence evaluate

$\displaystyle \int^{\frac{\pi}{2}}_{0} xsin^2xcos^2xdx$

After messing about with it I can never seem to get it into a nicer form... thanks any help would be appreciated (Yes)

Do you just want to evaluate the integral?

Try substituting $\displaystyle 1-cos^2(x)=sin^2(x)$ to get:

$\displaystyle \int^{\frac{\pi}{2}}_{0} x(1-cos^4x)dx$

I then suggest using tabular integration to solve that.
• Feb 6th 2008, 05:55 AM
Krizalid
Quote:

Originally Posted by slevvio
evaluate

$\displaystyle \int^{\frac{\pi}{2}}_{0} xsin^2xcos^2xdx$

Flip it around and substitute $\displaystyle u = \frac{\pi } {2} - x,$ $\displaystyle \varphi$ becomes (say the integral):

$\displaystyle \varphi = \int_0^{\pi /2} {\bigg( {\frac{\pi } {2} - x} \bigg)\sin ^2 x\cos ^2 x\,dx} .$

Hence $\displaystyle \varphi = \frac{\pi } {2}\int_0^{\pi /2} {\sin ^2 x\cos ^2 x\,dx} - \int_0^{\pi /2} {x\sin ^2 x\cos ^2 x\,dx} .$

The rest follows easily, the remaining integral can be computed without problems.