# Thread: Definite Integral problem

1. ## 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!

2. 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!
your integral is fine. I evaluate it as 0.8779... change your mode to radians.

3. 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.

4. Originally Posted by skeeter
your integral is fine. I evaluate it as 0.8779... change your mode to radians.
Skeeter is correct! Change your mode to radians. This completely explains the discrepancy!

5. Originally Posted by SammyS
Skeeter is correct! Change your mode to radians. This completely explains the discrepancy!
I did say "assuming you have inputted the terminals correctly"...

I guess the saying is true, to ASSUME makes an ASS of U and ME...