# Definite Integral problem

• Jan 30th 2011, 03:16 PM
Vamz
Definite Integral problem
$
\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\int\frac{1}{2}[1-\cos(8x)]\frac{1}{2}[1+\cos(8x)]\,dx
$

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

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

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

evaluates out to be:
$
\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!
• Jan 30th 2011, 03:29 PM
skeeter
Quote:

Originally Posted by Vamz
$
\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\int\frac{1}{2}[1-\cos(8x)]\frac{1}{2}[1+\cos(8x)]\,dx
$

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

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

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

evaluates out to be:
$
\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!

• Jan 30th 2011, 03:31 PM
Prove It
Let's see...

$\displaystyle \int{\sin^2{(4x)}\cos^2{(4x)}\,dx} = \int{[\sin{(4x)}\cos{(4x)}]^2\,dx}$

$\displaystyle = \int{\left[\frac{1}{2}\sin{(8x)}\right]^2\,dx}$

$\displaystyle = \int{\frac{1}{4}\sin^2{(8x)}\,dx}$

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

$\displaystyle = \int{\frac{1}{8}\left[1 - \cos{(16x)}\right]\,dx}$.

So I agree with your indefinite integral ( $\displaystyle +C$).

Assuming you have inputted your terminals correctly, you should be correct.
• Jan 30th 2011, 03:39 PM
SammyS
Quote:

Originally Posted by skeeter