# Intergrating sin^2 x cos^2 x

• Apr 8th 2008, 11:54 PM
Flay
Intergrating sin^2 x cos^2 x
This question has got me stumped;

Show that $\displaystyle \sin x \cos x = \frac{1}{2}\sin 2x$, and hence or otherwise find the exact value of $\displaystyle \int_0^\frac{\pi}{8} \sin^2x\cos^2x$.

The first half is easy, it's the integral that's giving me problems.
• Apr 9th 2008, 03:20 AM
Mathstud28
here is what you do
Quote:

Originally Posted by Flay
This question has got me stumped;

Show that $\displaystyle \sin x \cos x = \frac{1}{2}\sin 2x$, and hence or otherwise find the exact value of $\displaystyle \int_0^\frac{\pi}{8} \sin^2x\cos^2x$.

The first half is easy, it's the integral that's giving me problems.

rewrite $\displaystyle \int_0^{\frac{\pi}{8}}sin(x)^2\cdot{cos(x)^2}dx$ as $\displaystyle \int_0^{\frac{\pi}{8}}\bigg(\frac{1-cos(2x)}{2}\bigg)\cdot\bigg(\frac{1+cos(2x)}{2}\bi gg)$...and when you run into the obvious problem you will just look how I took care of it in the integral and apply the same concept except now you have to have $\displaystyle cos(2(2x))$
• Apr 9th 2008, 06:40 AM
mr fantastic
Quote:

Originally Posted by Mathstud28
rewrite $\displaystyle \int_0^{\frac{\pi}{8}}sin(x)^2\cdot{cos(x)^2}dx$ as $\displaystyle \int_0^{\frac{\pi}{8}}\bigg(\frac{1-cos(2x)}{2}\bigg)\cdot\bigg(\frac{1+cos(2x)}{2}\bi gg)$...and when you run into the obvious problem you will just look how I took care of it in the integral and apply the same concept except now you have to have $\displaystyle cos(2(2x))$

$\displaystyle \cos (2A) = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1$. Therefore:

$\displaystyle \cos^2 A = \frac{1}{2} (\cos (2A) + 1)$.

You'll need this to deal with the aforementioned obvious problem.