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$$\int \cos^4x\sin^4xdx$$

How should I approach this?

I know that $\sin^2x={1-\cos2x\over 2}$ and $\cos^2x={1+\cos2x\over 2}$

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    $\begingroup$ You could start with $\sin 2x =2\sin x \cos x$ $\endgroup$ – Mark Bennet Jan 9 '16 at 20:18
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    $\begingroup$ Given what you know, $$\cos^4(x) = (\cos^2(x))^2 = (1+2\cos(2x)+\cos^2(2x))/4.$$ Now, you can apply this again. $\endgroup$ – Mark McClure Jan 9 '16 at 20:18
  • $\begingroup$ @MarkBennet you mean to do $\frac{1}{16} \int (sinxcosx)^4=\frac{1}{16} \int (sin2x)^4$? $\endgroup$ – gbox Jan 9 '16 at 20:25
  • $\begingroup$ @gbox That gets the total exponent down from $8$ to $4$ in one simple move and keeps you with just one term. Then you can use the other identities you have with that. $\endgroup$ – Mark Bennet Jan 9 '16 at 20:33
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    $\begingroup$ @gbox I mean $4+4=8$ $\endgroup$ – Mark Bennet Jan 9 '16 at 20:38
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Continuing from where you left off:

$$\sin^2x\cos^2x=\left({1-\cos2x\over 2}\right)\left({1+\cos2x\over 2}\right)$$

$$\sin^2x\cos^2x=\frac{1}{4}(1 - \cos^2(2x))$$

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

Squaring both the sides:

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

This can be integrated in two ways:

Method 1: $$\sin^4x\cos^4x=\frac{1}{16}(\sin^4(2x))$$

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

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

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

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

$$=\frac{1}{32}[1 - \cos{(4x)}] - \frac{1}{128}[1 - \cos{(8x)}]$$ Which is easy to integrate

Method 2:

$$\:\frac{1}{16}\left[\sin^2(2x)\right]^2\;\;$$ $$=\frac{1}{16}\left[\frac{1\,-\,\cos(4x)}{2}\right]^2\:$$ $$=\:\frac{1}{64}\left[1\,-\,2\cdot\cos(4x) \,+\,\cos^2(4x)\right]$$ $$=\:\frac{1}{64}\left[1\,-\,2\cdot\cos(4x) \,+\,\frac{1+\cos(8x)}{2}\right]$$ $$=\:\frac{1}{128}\left[3\,-\,4\cdot\cos(4x)\,+\,\cos(8x)\right]$$

Which is again easy to integrate

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Use $$ \cos x=\frac{e^{ix}+e^{-ix}}{2}, \qquad \sin x=\frac{e^{ix}-e^{-ix}}{2i} $$ so \begin{align} \cos^4x\sin^4x &=\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\[6px] &=\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\[6px] &=\frac{1}{128}\cos8x-\frac{1}{32}\cos4x+\frac{3}{128} \end{align}

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we now that $$ \sin 2x=2\cos x\sin x$$ $$\cos x\sin x=\frac{\sin 2x}{2}$$ so... $$\cos^4 x\sin^4 x=\frac{\sin^4 2x}{16}$$ $$\frac{\sin^4 2x}{16}=\frac{(\frac{1-\cos 4x}{2})^2}{16}=\frac{1-2\cos 4x+\cos^24x}{64}=\frac{1-2\cos 4x+\frac{1+\cos 8x}{2}}{64}$$

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