# Alternate solutions to similar integrals?

So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integrals of this nature in general? I'd also like to know where I went wrong with this solution. \begin{align*} \int\sin^4x\,\mathrm{d}x&\rightarrow\\ u_1&=\sin^4x\\ u_1'&=4\cos x\sin^3x\\ v_1&=x\\ v_1'&=1\\ \leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-4\int x\cos x\sin^3x\,\mathrm{d}x\\ \rightarrow\int x\cos x\sin^3x\,\mathrm{d}x&\rightarrow\\ u_2&=\sin^3x\\ u_2'&=3\cos x\sin^2x\\ v_2&=\int x\cos x\,\mathrm{d}x\\ v_2'&=x\cos x\\ \rightarrow\int x\cos x\,\mathrm{d}x\rightarrow\\ u_3&=x\\ u_3'&=1\\ v_3&=\sin x\\ v_3'&=\cos x\\ \leftarrow\int x\cos x\,\mathrm{d}x&=x\sin x-\int\sin x\,\mathrm{d}x\\ &=x\sin x+\cos x\\ \leftarrow\int x\cos x\sin^3x\,\mathrm{d}x&=\left(\sin^3x\right)\left(x\sin x+\cos x\right)-\int\left(3\cos x\sin^2x\right)\left(x\sin x+\cos x\right)\,\mathrm{d}x\\ &=x\sin^4x+\cos x\sin^3x-3\int x\cos x\sin^3 x\,\mathrm{d}x+3\int\cos^2 x\sin^2x\,\mathrm{d}x\\ 4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4 x+\cos x\sin^3 x+3\int\cos^2x\sin^2x\,\mathrm{d}x\\ &=x\sin^4 x+\cos x\sin^3 x+3\int\left(1-\sin^2x\right)\sin^2x\,\mathrm{d}x\\ &=x\sin^4 x+\cos x\sin^3 x+3\int\sin^2x\,\mathrm{d}x-\int\sin^4x\,\mathrm{d}x\\ \rightarrow\int\sin^2x\,\mathrm{d}x\rightarrow\\ \cos2x&=1-2\sin^2x\\ \sin^2x&=\frac{1}{2}(1-\cos2x)\\ \leftarrow\frac{1}{2}\int1-\cos2x\,\mathrm{d}x&=\frac{1}{2}x-\frac{1}{4}\sin{2x}\\ \leftarrow4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4 x+\cos x\sin^3 x+3\left[\frac{1}{2}x-\frac{1}{4}\sin{2x}\right]-\int\sin^4x\,\mathrm{d}x\\ &=x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\\ \leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-\left[x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\right]\\ &=x\sin^4x-x\sin^4x-\cos x\sin^3x-\frac{3}{2}x+\frac{3}{4}\sin2x+\int\sin^4x\,\mathrm{d}x\\ \end{align*} As you can see, I can't continue because the integral cancels itself.

$\sin^2 x = \frac 12 (1 - \cos 2x)\\ \sin^4 x = \frac 14 (1 - \cos 2x)^2 = \frac 14 (1 - 2\cos 2x + \cos^2 2x)\\ \cos^2 2x = \frac 12 (1+\cos 4x)\\ \sin^4 x = \frac 14 (1 - \cos 2x)^2 = \frac 14 (1 - 2\cos 2x + \frac 12(1+\cos 4x))$

and simplify.

This is a nifty trick you can use if you know Euler's identity.

$e^{ix} = \cos x + i\sin x\\ \cos x = \frac {e^{ix} + e^{-ix}}{2}, \sin x = \frac {e^{ix} - e^{-ix}}{2i}\\ \sin^4 x = \frac {e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix}}{2^4}\\ \sin^4 x = \frac {\cos 4x + 4\cos 2x + 3}{8}\\$

Consider:$$\int\sin^4(x)\,\mathrm{d}x$$Apply the reduction formula:$$=\frac{3}{4}\cdot\int\sin^2(x)\,\mathrm{d}x\,-\frac{\cos(x)\sin^3(x)}{x}$$$$=\frac{3}{4}\cdot\left(\frac{1}{2}x-\frac{\cos(x)\sin(x)}{x}\right)-\frac{\cos(x)\sin^3(x)}{x}$$$$\boxed{=\frac{\sin(4x)-8\sin(2x)+12x}{32}+C}$$

Hint $$I=\int sin^4(x)dx=\int \sin^2(x)-\sin^2(x)\cos^2(x)dx$$ $$I=\int \sin^2(x)-\frac 14\sin^2(2x)dx$$

Note that $$\sin^2(x)=1-\cos^2(x)=1-\frac 12(\cos(2x)+1)=\frac 12(1 -\cos(2x))$$

$$I=\frac 12\int (1 -\cos(2x)) -\frac 18\int (1 -\cos(4x))dx$$ $$\boxed{I=\frac 38x-\frac 14 \sin(2x)+\frac 1{32} \sin(4x)+K}$$