<s>[INST] How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
 [/INST] $$\int_0^a f(x)\;dx=2\int_0^\dfrac a2 f(x)\;dx\; if \;f(a-x)=f(x)\;$$
and$\int_0^a f(x)\;dx=0\;\;if \;f(a-x)=-f(x)$
here $f(x)=\sin ^6x\implies f(\pi-x)=\sin ^6(\pi-x)\implies \sin^6x\implies f(x)$
so we can write it as first identity:
$$I=\int_0^\dfrac \pi2 \sin^6 x\;dx$$
from identity:$\int_0^af(x)\;dx=\int_0^af{(a-x)}\;dx$
$$I=\int_0^\dfrac \pi2 \cos^6 x\;dx$$
$$2I=\int_0^\dfrac \pi2 \sin^6 x+\cos ^6 x\;dx$$
$$2I=\int_0^\dfrac \pi2 ({\sin^2 x})^3+({\cos ^2 x})^3\;dx$$
$$2I=\int_0^\dfrac \pi2 (\sin^2 x+\cos ^2 x)({\sin^4 x}+{\cos ^4 x}-\sin^2 x\cos^2 x)\;dx$$
$$2I=\int_0^\dfrac \pi2 ({\sin^4 x}+{\cos ^4 x}+2\sin^2 x\cos^2 x-3\sin^2 x\cos^2 x)\;dx$$
$$2I=\int_0^\dfrac \pi2 ({{\sin^2 x}+{\cos ^2 x}}^2)-3\sin^2 x\cos^2 x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-3\sin^2 x\cos^2 x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34{(2\sin x\cos x)}^2\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34\sin^2 2x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34\dfrac{(1-\cos 4x)}2\;dx$$
$$2I=\left[x-\dfrac 38(x-\dfrac{\sin 4x}4)\right]_0^\dfrac \pi2 \implies \left[\dfrac {5x}{8}+\dfrac{3\sin 4x}{32}\right]_0^\dfrac \pi2 \implies \left[\dfrac {5\pi}{16}\right]\implies  I=\left[\dfrac {5\pi}{32}\right]$$
proof of identity:$$\int_0^{2a} f(x)\;dx=2\int_0^a f(x)\;dx\; if \;f(a-x)=f(x)\;$$
this identity can be prove using this identity:$$\int _0^{2a} f(x)=\int _0^a f(x)\;dx+\int_0^a f(2a-x)\;dx$$
when $f(2a-x)=f(x)$ this become
$$\int _0^{2a} f(x)=\int _0^a f(x)\;dx+\int_0^a f(x)\;dx$$
$$\int _0^{2a} f(x)=2\int_0^a f(x)\;dx$$
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