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I'm having some trouble figuring out the right substitutions to make to integrate

$$\int \sin^4(\theta)d\theta$$

and

$$\int \cos^4(\theta)d\theta$$

Any hints or suggestions are welcome.

Thanks,

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    $\begingroup$ You could always start with $\sin^4(\theta) = \sin^2(\theta)(1 - \cos^2(\theta)) = \sin^2(\theta) - \sin^2(\theta)\cos^2\theta$. Then maybe you know how to proceed using double angle formulas? $\endgroup$ – froggie Nov 9 '15 at 3:30
  • $\begingroup$ Ah, that's where I was stuck -- ok, I will proceed from here with the double-angle formula. Thanks for your quick response @froggie :-) $\endgroup$ – User001 Nov 9 '15 at 3:32
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    $\begingroup$ You're welcome, Lebron! $\endgroup$ – froggie Nov 9 '15 at 3:33
  • $\begingroup$ Hi @froggie, now I am stuck with the integrand $sin^2(\theta) - \frac{sin^2(2\theta)}{4}$. Where can I go from here? $\endgroup$ – User001 Nov 9 '15 at 3:42
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    $\begingroup$ I'm guessing the trouble is with the $\sin^2\theta$ term? This can be rewritten again using a double angle identity, but now for cosine: $\cos(2\theta) = \cos^2\theta - \sin^2\theta = (1 - \sin^2\theta) - \sin^2\theta$. Solving for $\sin^2\theta$ gives $\sin^2\theta = (1 - \cos(2\theta))/2$. $\endgroup$ – froggie Nov 9 '15 at 3:44
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Notice that you can write $\sin^4(x)$ as follows:

\begin{align} \sin^4(x) = &\ (\sin^2(x))^2 = \Big(\frac{1 - \cos(2x)}{2}\Big)^2 \\ = &\ \frac{1 -2\cos(2x) + \cos^2(2x)}{4} \\ = &\ \frac{1}{4} - \frac{\cos(2x)}{2} + \frac{1 + \cos(4x)}{8}. \end{align}

For $\cos^4(x)$ we procede as above:

\begin{align} \cos^4(x) = &\ (\cos^2(x))^2 = \Big(\frac{1 + \cos(2x)}{2}\Big)^2 \\ = &\ \frac{1 + 2\cos(2x) + \cos^2(2x)}{4} \\ = &\ \frac{1}{4} + \frac{\cos(2x)}{2} + \frac{1 + \cos(4x)}{8}. \end{align}

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HINT (using partial integration):

$$\int\sin^4(x)\space\space\text{d}x=$$ $$-\frac{1}{4}\sin^3(x)\cos(x)+\frac{3}{4}\int\sin^2(x)\space\space\text{d}x=$$ $$-\frac{1}{4}\sin^3(x)\cos(x)+\frac{3}{4}\int\left(\frac{1}{2}-\frac{1}{2}\cos(2x)\right)\space\space\text{d}x=$$ $$-\frac{1}{4}\sin^3(x)\cos(x)-\frac{3}{8}\int\cos(2x)\space\space\text{d}x+\frac{3}{8}\int 1\space\space\text{d}x$$


$$\int\cos^4(x)\space\space\text{d}x=$$ $$\frac{1}{4}\sin(x)\cos^3(x)+\frac{3}{4}\int\cos^2(x)\space\space\text{d}x=$$ $$\frac{1}{4}\sin(x)\cos^3(x)+\frac{3}{4}\int\left(\frac{1}{2}+\frac{1}{2}\cos(2x)\right)\space\space\text{d}x=$$ $$\frac{1}{4}\sin(x)\cos^3(x)+\frac{3}{8}\int\cos(2x)\space\space\text{d}x+\frac{3}{8}\int 1\space\space\text{d}x$$

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