annoying integral problem so close yet so far

Dec 2012
128
1
NY
Ok the original problem is

Integral sin^(4) x cos^(4) x

After making the two of those sin^2 and and cos^2 and doing trig function substation my integral became

1/4 integral [1-cos^2 2x]^2

Then 1/4 [sin^2 2x]^2
I am officially I stuck and I know I'm close to solving it but I am now waving the white flag

Please help!
 
Nov 2009
354
18
\(\displaystyle \begin{align*}\int \sin^{4}(x) \cos^{4} (x)\,\,dx =& \int \frac{(2\sin(x)\cos(x))^{4}}{16}\,\,dx\\=& \int \frac{(\sin(2x))^{4}}{16}\,\,dx....\text{[}\sin(2x) = 2\sin(x)\cos(x)\text{]}\\=& \int \frac{(\sin^{2}(2x))^{2}}{16}\,\,dx\\=&\int \frac{\left(1 - \cos^{2}(2x)\right)^2}{16}\,\,dx......\text{[}\sin^{2}(2x) = 1 - \cos^{2}(2x)\text{]}\\ =& \frac{1}{16} \int \left(\frac{1}{2} - \frac{\cos(4x)}{2}\right)^2\,\,dx....\text{[}\cos(4x) = 2\cos^{2}(2x) -1\text{]}\\=& \frac{1}{16} \int \left( \frac{1}{4} - \frac{\cos(4x)}{2} + \frac{\cos^{2}(4x)}{4} \right)\,\,dx\\=& \frac{1}{16} \int \left( \frac{3}{8} - \frac{\cos(4x)}{2} + \frac{\cos(8x)}{8}\right)\,\,dx....\text{[}\cos(8x) = 2\cos^{2}(4x) -1\text{]} \\=& \frac{1}{16} \left( \frac{3x}{8} - \frac{\sin(4x)}{8} + \frac{\sin(8x)}{64}\right) + C \\=& \frac{3x}{128} -\frac{\sin(4x)}{128} + \frac{\sin(8x)}{1024} + C....\text{[Answer]}\end{align*}\)

You can check the answer here:

int (sin(x))^4 (cos(x))^4 - Wolfram|Alpha
 
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