$$\int \cos^4x\sin^4xdx$$
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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3You could start with $\sin 2x =2\sin x \cos x$ – Mark Bennet Jan 09 '16 at 20:18
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2Given what you know, $$\cos^4(x) = (\cos^2(x))^2 = (1+2\cos(2x)+\cos^2(2x))/4.$$ Now, you can apply this again. – Mark McClure Jan 09 '16 at 20:18
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@MarkBennet you mean to do $\frac{1}{16} \int (sinxcosx)^4=\frac{1}{16} \int (sin2x)^4$? – gbox Jan 09 '16 at 20:25
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@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. – Mark Bennet Jan 09 '16 at 20:33
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@MarkBennet I do not get it, I switch between two expression with exponent of 4 – gbox Jan 09 '16 at 20:37
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1@gbox I mean $4+4=8$ – Mark Bennet Jan 09 '16 at 20:38
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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}
egreg
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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
Ashish Gupta
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we now that $$ \sin 2x=2\cos x\sin x$$ $$\cos x\sin x=\frac{\sin 2x}{2}$$ so... $$\cos^4 x\sin^4 x=\frac{\sin^4 2x}{16}$$ $$\frac{\sin^4 2x}{16}=\frac{(\frac{1-\cos 4x}{2})^2}{16}=\frac{1-2\cos 4x+\cos^24x}{64}=\frac{1-2\cos 4x+\frac{1+\cos 8x}{2}}{64}$$
E.H.E
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