Integral

**slevvio** · February 6th 2008, 03:52 AM

Hello, some of you may remember me from the previous question.

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

Hence evaluate

$ \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

**galactus** · February 6th 2008, 04:55 AM

You could try rewriting your integral.

$xsin^{2}xcos^{2}x=\frac{x^{2}}{8}-\frac{x^{2}cos(4x)}{8}$

**colby2152** · February 6th 2008, 05:24 AM

Originally Posted by slevvio

Hello, some of you may remember me from the previous question.

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

Hence evaluate

$ \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

Do you just want to evaluate the integral?

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

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

I then suggest using tabular integration to solve that.

**Krizalid** · February 6th 2008, 05:55 AM

Originally Posted by slevvio

evaluate

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

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

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

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

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

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