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,
$\endgroup$ 52 Answers
$\begingroup$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}
$\endgroup$ $\begingroup$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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