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(\frac{1-\cos(2x)}2\right)^k`

Expand the binomial, then use the half-angle identity

`\cos^2(x)=\frac{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.