# Integration!

• December 4th 2012, 09:17 AM
Tutu
Integration!
How do I integrate sin^4(2x) ?
Note: The 4 is the power but the 2x is not.

I tried and I got 3/8x - (1/8)sin4x + (1/64)sin8x + c but the answer is 1/32(48x-4sin4x+sin8x)+c..
• December 4th 2012, 09:42 AM
topsquark
Re: Integration!
Quote:

Originally Posted by Tutu
How do I integrate sin^4(2x) ?
Note: The 4 is the power but the 2x is not.

I tried and I got 3/8x - (1/8)sin4x + (1/64)sin8x + c but the answer is 1/32(48x-4sin4x+sin8x)+c..

Here's a quick run-down.

$\int sin^4(2x)~dx$

First a variable change. Let y = 2x, etc
$= \frac{1}{2} \int sin^4(y)~dy$

Clever trick warning!!

$= \frac{1}{2} \int \left ( sin^2(y) \right ) ^2dy$

And from $cos(2 \theta ) = 1 - 2~sin^2( \theta )$. Plugging this in:

$= \frac{1}{2} \int \left ( \frac{1}{2}(1 - cos(2y) \right ) ^2dy$

$= \frac{1}{8} \int (1 - 2~cos(2y) + cos^2(2y))dy$

See what you can do from here.

-Dan
• December 4th 2012, 04:26 PM
Tutu
Re: Integration!
When I let y=2x why do.I take the 1/2 out? Is it because cause 2x=2sinxcosx?.thanks!
• December 4th 2012, 04:35 PM
sjmiller
Re: Integration!
Quote:

Originally Posted by Tutu
When I let y=2x why do.I take the 1/2 out? Is it because cause 2x=2sinxcosx?.thanks!

It is the subsitution rule for integration. dy=2dx --> dy/2=dx
Normally you will see the variable 'u' used instead of 'y'... but which you use doesn't matter.
• December 4th 2012, 05:23 PM
Soroban
Re: Integration!
Hello, Tutu!

Quote:

$\int \sin^4(2x)\,dx$

I tried and I got: . $\tfrac{3}{8}x - \tfrac{1}{8}\sin4x + \tfrac{1}{64}\sin8x + C$ . You are right!

but the answer is: . $\tfrac{1}{32}(48x-4\sin4x+\sin8x)+C$ . They are wrong!

Here are my steps:

$\int\sin^4(2x)\,dx \;=\;\int\left[\sin^2(2x)\right]^2dx$

. . . . . . . . . . $=\;\int\left[\frac{1-\cos4x}{2}\right]^2dx$

. . . . . . . . . . $=\;\tfrac{1}{4}\int\big[1 - 2\cos(4x) + \cos^2(4x)\big]\,dx$

. . . . . . . . . . $=\;\tfrac{1}{4}\int\left[1 - 2\cos(4x) + \frac{1 + \cos(8x)}{2}\right]\,dx$

. . . . . . . . . . $=\;\tfrac{1}{4}\int \left[\tfrac{3}{2} - 2\cos(4x) + \tfrac{1}{2}\cos(8x)\right]\,dx$

. . . . . . . . . . $=\;\tfrac{1}{4}\left[\tfrac{3}{2}x - \tfrac{1}{2}\sin(4x) + \tfrac{1}{16}\son(8x)\right] + C$

. . . . . . . . . . $=\;\tfrac{3}{8}x - \tfrac{1}{8}\sin(4x) + \tfrac{1}{64}\sin(8x) + C$