Question

\int\limits^0_\pi {x*sin^{m} (x)} \, dx

Answer 1

Let

$I(m) = \displaystyle \int_0^\pi x\sin^m(x)\,\mathrm dx$

Integrate by parts, taking

u = x   ==>   du = dx

dv = sinᵐ (x) dx   ==>   v = ∫ sinᵐ (x) dx

so that

$I(m) = \displaystyle uv\bigg|_{x=0}^{x=\pi} - \int_0^\pi v\,\mathrm du = -\int_0^\pi \sin^m(x)\,\mathrm dx$

There is a well-known power reduction formula for this integral. If you want to derive it for yourself, consider the cases where m is even or where m is odd.

If m is even, then m = 2k for some integer k, and we have

$\sin^m(x) = \sin^{2k}(x) = \left(\sin^2(x)\right)^k = \left(\dfrac{1-\cos(2x)}2\right)^k$

Expand the binomial, then use the half-angle identity

$\cos^2(x)=\dfrac{1+\cos(2x)}2$

as needed. The resulting integral can get messy for large m (or k).

If m is odd, then m = 2k + 1 for some integer k, and so

$\sin^m(x) = \sin(x)\sin^{2k}(x) = \sin(x)\left(\sin^2(x)\right)^k = \sin(x)\left(1-\cos^2(x)\right)^k$

and then substitute u = cos(x) and du = -sin(x) dx, so that

$I(2k+1) = \displaystyle -\int_0^\pi\sin(x)\left(1-\cos^2(x)\right)^k = \int_1^{-1}(1-u^2)^k\,\mathrm du = -\int_{-1}^1(1-u^2)^k\,\mathrm du$

Expand the binomial, and so on.