integral of sin^4 (x) * cos^4 (x)

[math]

integral of sin^4 (x) * cos^4 (x)
I tried a bunch of u substitutions but nothing so far has worked. I know that sin^2 x + cos^2 x = 1 and that cos2x = 2cos^2 (x) - 1 = 1-2sin^2 (x) and I can solve either of these for sin^2 (x) or cos^2 (x). Will any of this help me? Thanks.

 

[skeeter]

half-angle formulas ... a few times

\(\displaystyle \L \sin^4{x} \cdot \cos^4{x} =\)

\(\displaystyle \L (\sin^2{x})^2(\cos^2{x})^2 =\)

\(\displaystyle \L \left(\frac{1 - \cos{(2x)}}{2}\right)^2\left(\frac{1 + \cos{(2x)}}{2}\right)^2 =\)

\(\displaystyle \L \frac{1}{16} \left[1 - \cos^2{(2x)}\right]^2 =\)

\(\displaystyle \L \frac{1}{16} \sin^4{(2x)} =\)

\(\displaystyle \L \frac{1}{16} \sin^2{(2x)} (1 - \cos^2{(2x)}) =\)

\(\displaystyle \L \frac{1}{16} \left[\sin^2{(2x)} - \sin^2{(2x)} \cos^2{(2x)}\right] =\)

\(\displaystyle \L \frac{1}{16} \left[\sin^2{(2x)} - \left(\frac{1-\cos{(4x)}}{2}\right)\left(\frac{1+\cos{(4x)}}{2}\right)\right] =\)

\(\displaystyle \L \frac{1}{16} \left[\sin^2{(2x)} - \frac{1}{4}(1 - \cos^2{(4x)}\right] =\)

\(\displaystyle \L \frac{1}{16} \left[\sin^2{(2x)} - \frac{1}{4}\sin^2{(4x)}\right] =\)

\(\displaystyle \L \frac{1}{32}[1 - \cos{(4x)}] - \frac{1}{128}[1 - \cos{(8x)}]\)

the last expression should be relatively simple to integrate.

 

[math]

thanks skeeter!

 

[soroban]

Hello, math!

A slightly different approach . . .

\(\displaystyle \L\int \sin^4x\cdot\cos^4x\,dx\)

We have: \(\displaystyle \L\\sin x\cdot\cos x)^4 \:=\:\left(\frac{2\cdot\sin x\cdot\cos x}{2}\right)^4 \:=\:\left[\frac{\sin(2x)}{2}\right]^4\:=\:\frac{1}{16}\cdot\sin^4(2x)\)

. . \(\displaystyle \L= \:\frac{1}{16}\cdot\left[\sin^2(2x)\right]^2\;=\;\frac{1}{16}\cdot\left[\frac{1\,-\,\cos(4x)}{2}\right]^2\:=\:\frac{1}{64}\cdot\left[1\,-\,2\cdot\cos(4x) \,+\,cos^2(4x)\right]\)

. . \(\displaystyle \L= \:\frac{1}{64}\cdot\left[1\,-\,2\cdot\cos(4x)\,+\,\frac{1\,+\,\cos(8x)}{2}\right] \:=\:\frac{1}{128}\cdot\left[3\,-\,4\cdot\cos(4x)\,+\,\cos(8x)\right]\)


And we have: \(\displaystyle \L\;\frac{1}{128}\int\left[3\,-\,4\cdot\cos(4x)\,+\,\cos(8x)\right]\,dx\)

. . Go for it!