Integrate $\sin^4(x)$

Question

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 $\frac{3}{8}x -\frac{1}{4}\sin(2x)+\frac{1}{32}\sin(4x)+C$.

Answer 1

First of all, I think you mistyped the last line of your integral, changing a 2 for an 8. Besides, you can use the linearity of the integral to write $$ \int(1-2\cos(2x)+\cos^{2}(2x))dx = \int 1 dx - 2\int \cos(2x)dx+\int \cos^{2}(2x)dx. $$ Now, note that $$\cos^{2}(2x) + \sin^{2}(2x) = 1$$ and $$ \cos^{2}(2x)-\sin^{2}(2x) = \cos(4x)$$ so that $$\cos^{2}(2x) = \frac{1+\cos(4x)}{2}.$$ Thus, we have $$ \int \cos^{2}(2x) = \int \frac{1+\cos(4x)}{2}dx = \frac{1}{2}x + \frac{\sin(4x)}{8}+C$$. Now, the previous integrals are given by $$\int 1 dx -2\int \cos(2x) = x -2\frac{\sin(2x)}{2}+C = x - \sin(2x) +C.$$

Finally, your integral is $$\frac{1}{4}[x-\sin(2x)+\frac{1}{2}x+\frac{\sin(4x)}{8}]+C = \frac{3}{8}x-\frac{1}{4}\sin(2x)+\frac{\sin(4x)}{32} + C$$

Answer 2

Surely your $8$ should be a $2$ (not to mention each $\cos^2x$ should be $\cos^22x$), giving$$\frac14\int(1-2\cos 2x+\cos^22x)dx=\frac18\int(3-4\cos 2x+\cos 4x)dx=\frac{1}{32}(12x-8\sin 2x+\sin 4x)+C.$$