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.
$\displaystyle \int sin^4(2x)~dx$
First a variable change. Let y = 2x, etc
$\displaystyle = \frac{1}{2} \int sin^4(y)~dy$
Clever trick warning!!
$\displaystyle = \frac{1}{2} \int \left ( sin^2(y) \right ) ^2dy$
And from $\displaystyle cos(2 \theta ) = 1 - 2~sin^2( \theta )$. Plugging this in:
$\displaystyle = \frac{1}{2} \int \left ( \frac{1}{2}(1 - cos(2y) \right ) ^2dy$
$\displaystyle = \frac{1}{8} \int (1 - 2~cos(2y) + cos^2(2y))dy$
See what you can do from here.
-Dan
Hello, Tutu!
Here are my steps:$\displaystyle \int \sin^4(2x)\,dx$
I tried and I got: .$\displaystyle \tfrac{3}{8}x - \tfrac{1}{8}\sin4x + \tfrac{1}{64}\sin8x + C$ . You are right!
but the answer is: .$\displaystyle \tfrac{1}{32}(48x-4\sin4x+\sin8x)+C$ . They are wrong!
$\displaystyle \int\sin^4(2x)\,dx \;=\;\int\left[\sin^2(2x)\right]^2dx$
. . . . . . . . . . $\displaystyle =\;\int\left[\frac{1-\cos4x}{2}\right]^2dx$
. . . . . . . . . . $\displaystyle =\;\tfrac{1}{4}\int\big[1 - 2\cos(4x) + \cos^2(4x)\big]\,dx$
. . . . . . . . . . $\displaystyle =\;\tfrac{1}{4}\int\left[1 - 2\cos(4x) + \frac{1 + \cos(8x)}{2}\right]\,dx$
. . . . . . . . . . $\displaystyle =\;\tfrac{1}{4}\int \left[\tfrac{3}{2} - 2\cos(4x) + \tfrac{1}{2}\cos(8x)\right]\,dx$
. . . . . . . . . . $\displaystyle =\;\tfrac{1}{4}\left[\tfrac{3}{2}x - \tfrac{1}{2}\sin(4x) + \tfrac{1}{16}\son(8x)\right] + C$
. . . . . . . . . . $\displaystyle =\;\tfrac{3}{8}x - \tfrac{1}{8}\sin(4x) + \tfrac{1}{64}\sin(8x) + C$