Note that $n$ is an arbitrary constant. $$ \int(\sin^n(x))dx $$ I start by using the obvious integrating by parts and get: $$ \frac{d}{dx}[x\sin^n(x)] = \sin^n(x) + nx\sin^{n-1}(x)\cos(x) $$
$$ x\sin(x) = \int(\sin^n(x))dx + n \int(x\sin^{n-1}(x)\cos(x))dx $$
$$ \int(\sin^n(x))dx = x\sin(x) - n \int(x\sin^{n-1}(x)\cos(x))dx $$ I tried to integrate using the knowledge I know but got nowhere fast, I checked wolframalpha expecting it to be something obvious but got $$ n \int(x\sin^{n-1}(x)\cos(x))dx \\ = n\left(\frac{\sin^{n}(x)(x+\cot(x)\cdot{}_2F_{1}\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos^{2}(x)\right)\sin^2(x)^{\frac{1}{2} - \frac{n}{2}}}{n}\right)+c $$
which I simplified down to $$ = x\sin^n(x) + \sin^{n-1}(x) \sin^{1-n}(x)\cos(x) \cdot {}_2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos^2(x)\right)+c $$
$$ =x\sin^n(x) + \cos(x)\cdot {}_2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos^2(x)\right) +c $$ But what I am oblivious to is the process that WolframAlpha took to get the integral of $x\sin^{n-1}(x)\cos(x)$. What steps were taken and where do I go to learn these steps?
That's OK, because there is nothing either helpful or spectacular about that process. Basically, he interprets the sine function as $\sqrt{1-\cos^2x}$, and expands the integrand using the binomial series, then switches the order of summation and integration, and reparses the whole expression in terms of hypergeometric functions. But this is merely a question of changing one notation into another; it does not help at all in establishing meaningful results, relations, and identities involving the initial integral. What is helpful, however, is understanding the fact that all natural powers of the sine and cosine functions can be rewritten as a linear combination of terms of the form $\sin kx$ and $\cos px$, whose integration then becomes trivial.