$\displaystyle

\displaystyle\int_{6}^{13} \sin^2( 4 x ) \cos^2 (4 x) \,dx

$

Can someone please point out where I went wrong in my work?

Thank you

First, I rewrote this problem as:

$\displaystyle

\displaystyle\int\frac{1}{2}[1-\cos(8x)]\frac{1}{2}[1+\cos(8x)]\,dx

$

$\displaystyle

\displaystyle\frac{1}{4}\int(1-\cos(8x))(1+\cos(8x))\,dx = \frac{1}{4}\int1-\cos^2(8x)\,dx

$

$\displaystyle

\displaystyle\frac{1}{4}\int1-\frac{1}{2}[1+\cos(16x)]\,dx=\frac{1}{4}\int\frac{1-cos(16x)}{2}\,dx

$

$\displaystyle

\displaystyle\frac{1}{8}\int1-\cos(16x)\,dx

$

evaluates out to be:

$\displaystyle

\displaystyle\frac{x}{8}-\frac{\sin(16x)}{128}

$

Now, when I plug in the upper & lower limits, I get:

.886437448892

as a final answer. This does not appear to be correct. What am I doing wrong?

Thanks!