Question

Solve : $I_n={\int }_0^{\frac{\pi }{2}}{e}^{-x}{\sin }^{n}xdx$

Answer 1

$I_n={\int }_0^{\frac{\pi }{2}}{e}^{-x}{\sin }^{n}xdx$

show that $I_n=-e^{\frac{-\pi }{2}}+n{\int }_0^{\frac{\pi }{2}}{e}^{-x}{\sin }^{n-1}xdx$

$I_n={\int }_0^{\frac{\pi }{2}}{e}^{-x}{\sin }^{n}xdx$

substitute $u={\sin }^{n}x,\frac{dv}{dx}=e^{-x}\Rightarrow v=-e^{-x}$

$I_n=[-e^{-x}{\sin }^{n}x{]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}-e^{-x}n{\sin }^{n-1}x\cos xdx$

$I_n=-e^{\frac{-\pi }{2}}+n{\int }_0^{\frac{\pi }{2}}-e^{-x}\cos x{\sin }^{n-1}xdx$

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${\int }_0^{\frac{\pi }{2}}{e}^{-x}\cos x{\sin }^{n-1}xdx$

$=[-e^{-x}\cos x{\sin }^{n-1}x{]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}{e}^{-x}((n-1){\sin }^{n-2}x-n{\sin }^{n}x)dx$

$=(n-1){\int }_0^{\frac{\pi }{2}}{e}^{-x}{\sin }^{n-2}xdx-n{\int }_0^{\frac{\pi }{2}}{e}^{-x}{\sin }^{n}xdx$

$=(n-1)I_{n-2}-n{I}_n$

$\therefore I_n=-e^{\frac{-\pi }{2}}+n(n-1)I_{n-2}-n^2{I}_n$

$\Rightarrow (n^2+1)I_n=-e^{\frac{-\pi }{2}}+n(n-1)I_{n-2}$