<s>[INST] Integrate $\sin^4(x)$ Consider the integral 
$$\int \sin^4(x)dx$$
Now I could separate the $\sin^4(x)$ into two $\sin^2(x)$ terms and the use power reducing formula
$$\int \sin^2(x)\sin^2(x)dx $$
$$\int \frac{1-\cos(2x)}{2}*\frac{1-\cos(2x)}{2}dx $$
$$\int \frac{(1-\cos(2x))^2}{4} dx$$
$$\int\frac{1}{4}-\frac{2\cos(2x)}{4}+\frac{\cos^2(x)}{4}dx $$
$$\frac{1}{4} \int1-8\cos(2x)+\cos^2(2x)dx $$
Now would it be good to rewrite that $\cos^2(2x)$ as $1-\sin^2(2x)$ and then integrate or should I also separate the terms into their own integrands?
The answer I am looking for which is $\frac{3}{8}x -\frac{1}{4}\sin(2x)+\frac{1}{32}\sin(4x)+C$.
 [/INST] Surely your $8$ should be a $2$ (not to mention each $\cos^2x$ should be $\cos^22x$), giving$$\frac14\int(1-2\cos 2x+\cos^22x)dx=\frac18\int(3-4\cos 2x+\cos 4x)dx=\frac{1}{32}(12x-8\sin 2x+\sin 4x)+C.$$
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