I am trying to solve this integral
$$\int_{0}^{\pi /2}\sin^n x\cdot dx.$$
I think we should solve it for:
a) odd numbers $2n+1$
$$\int_0^{\pi /2}\sin^{2n+1}x\cdot dx = \int_0^{\pi /2}\sin x\cdot \sin^{2n}x\cdot dx=\int_0^{\pi /2}\sin x\cdot (1-cos^2x) ^n\cdot dx$$
let $t=\cos(x)$ and $dt=-\sin(x) \, dx$ then:
$$\int_0^{\pi /2}\sin x\cdot (1-\cos^2x) ^n\cdot dx=-\int_1^0 (1-t^2)^n \, dt$$
Unfortunately I can not solve this integral. Would you please help me to finish it?
b) even numbers $2n$:
$$\int_0^{\pi /2}\sin^{2n}x\cdot dx = \int_0^{\pi /2}\left( \frac {1-\cos 2x} 2\right)^n \, dx$$
Unfortunately I can not solve this integral. Would you please help me to finish it?
I tried to search for something useful on the Internet and I found these two formulas: $$\int_0^{\pi /2}\sin^{2n+1}x\cdot dx = \int_0^{\pi /2}\cos^{2n+1}x\cdot dx = \frac{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}{1\cdot 3\cdot 5\cdot \ldots \cdot (2n+1)}\frac{\pi}{2}$$ $$\int_0^{\pi /2}\sin^{2n}x\cdot dx = \int_0^{\pi /2}\cos^{2n}x\cdot dx = \frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \ldots \cdot 2n}\frac{\pi}{2}$$
If you could write proofs of these two formulas that would solve my problem.
Thank you
You can do this by integration by parts, or equivalently using the product formula for differentiation. Let $I_n$ be the integral involving $\sin^n x$
So note that $$\frac {d}{dx}(\cos x\sin^{n-1}x)=-\sin^n x+(n-1)\cos^2x\sin^{n-2}x$$$$= -\sin^n x+(n-1)(1-\sin^2x)\sin^{n-2}x=-n\sin^nx+(n-1)sin^{n-2}x$$
Now integrate both sides, noting that: $\cos x\sin^{n-1}x=0$ at the limits of integration to obtain $$0=-nI_n+(n-1)I_{n-2}$$ which becomes $$I_n=\frac {(n-1)}nI_{n-2}$$
Using this successively gives the products you see in your answers, and takes you down to evaluating $I_1$ or $I_0$ (depending whether $n$ is odd or even). You should be able to finish it from there.