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

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

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.

2. half-angle formulas ... a few times

$\L \sin^4{x} \cdot \cos^4{x} =$

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

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

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

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

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

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

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

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

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

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

the last expression should be relatively simple to integrate.

3. thanks skeeter!

4. Re: integral of sin^4 (x) * cos^4 (x)

Hello, math!

A slightly different approach . . .

$\L\int \sin^4x\cdot\cos^4x\,dx$

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

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

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

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

. . Go for it!

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