Answer to Question #345841 in Calculus for Roy

Question

 if I_n = ₀∫∞ e^-xsinⁿ(x)dx , prove that (1+n²)I_n = n(n-1) I_n-2 for n >₌ 2

Answer 1

$I_n=\int\limits_0^\infty e^{-x}\sin^nx dx \\ =\int\limits_0^\infty (-e^{-x})^\prime \sin^nx dx \\=-e^{-x}\sin^nx\big|_0^\infty+\int\limits_0^\infty e^{-x}\cdot n\sin^{n-1}x\cos xdx \\ =0+n \int\limits_0^\infty (-e^{-x})^\prime \cdot\sin^{n-1}x\cos xdx \\ =-ne^{-x}\sin^{n-1}x\cos xdx \big|_0^\infty+n\int\limits_0^\infty e^{-x}\bigg( (n-1)\sin^{n-2}x\cos^2 x-\sin^n xdx\bigg)dx \\ =0+n(n-1)\int\limits_0^\infty e^{-x}\sin^{n-2}(1-\sin^2x)dx-n\int\limits_0^\infty e^{-x}\sin^nxdx \\ =n(n-1)I_{n-2}-n(n-1)I_n-nI_n$

$I_n=n(n-1)I_{n-2}-n(n-1)I_n-nI_n$

$(n^2+1)I_n=n(n-1)I_{n-2}$