# Integral Question

• Jun 4th 2010, 03:32 PM
Paymemoney
Integral Question
Hi

Can someone tell me why this is incorrect.

$\int sin^2(x)cos^2(x)dx$

$\int (0.5(1-cos(2x))(0.5(cos(2x)+1))dx$

$\int (0.5-\frac{cos(2x)}{2})(\frac{cos(2x)}{2}+0.5))dx$

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

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

$u=cos(2x)$
$du=-sin(2x)$

$\frac{x}{4} - \frac{1}{4} \int u^2 du$

$\frac{x}{4} - \frac{1}{4} * \frac{u^3}{3} +c$

$\frac{x}{4} - \frac{1}{4} * \frac{cos^3(2x)}{12} +c$

P.S
• Jun 4th 2010, 03:50 PM
ojones
For one thing, the substitution is incorrect. You have no $\sin (2x)$ in the integral to get to $u^2 du$. You have also differentiated $u$ incorrectly.
• Jun 4th 2010, 05:26 PM
sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...db8dce38-1.gif

Till the above step you are right.

Next write $cos^2(2x) = \frac{1}{2}(1+cos4x)$ and proceed.