Integrals reductive forms

Question

I have the integral: $$I_ν=\int_0^\pi e^{-x}(\sin{x})^ν\text{d}x$$ and I want to prove that: $$(ν^2+1)I_ν=ν(ν-1)I_{ν-2},ν\ge2$$ So, I am thinking of using the integration by parts method and I have: $$I_ν=\int_0^\pi e^{-x}(\sin{x})^ν\text{d}x=\int_0^\pi -(e^{-x})'(\sin{x})^ν\text{d}x=-\int_0^\pi (e^{-x})'(\sin{x})^ν\text{d}x=-[e^{-x}(\sin{x})^ν]_0^\pi +ν\int_0^\pi e^{-x}(\sin{x})^{ν-1}(\sin{x})'\text{d}x$$ But now I don't know how to continue because I will get a $\cos{x}$ inside my integral. Any ideas?

Answer 1

You can apply the same idea again:\begin{align}\int_0^\pi e^{-x}\sin^{\nu-1}(x)\cos(x)\,\mathrm dx&=\int_0^\pi e^{-x}\bigl((\nu-1)\sin^{\nu-2}(x)\cos^2(x)-\sin^\nu(x)\bigr)\,\mathrm dx\\&=\int_0^\pi e^{-x}\bigl((\nu-1)\sin^{\nu-2}(x)\bigl(1-\sin^2(x)\bigr)-\sin^\nu(x)\bigr)\,\mathrm dx.\end{align}Can you take it from here?