How can the recursive formula for the integral of sin^n(x)*e^-x be found?

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Homework Statement



Find a recursive formula for

I_{n}:=\int_{0}^{\infty}\sin^{n}(x)\cdot e^{-x}\ dx

Homework Equations





The Attempt at a Solution



I wrote I_{n+1} and tried to integrate by parts and with substitution, however, I wasn't able to get a the original term so that I could write it recursively.
 
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You need to do integration by parts twice and then remember that to pick the trig terms as 'u' when integrating by parts the second time
 
I did that and got something that is partially recursive

I_{n}=n^{2}\cdot\int_{0}^{\infty}\sin^{n-2}(x)\cdot\cos^{2}(x)\cdot e^{-x}\ dx-n\cdot I_{n-2}

However, I don't know how to eliminate the cos^2 at all. Moreover, one should prove with the recursive function that when n->infinity I_n->0. But with this result I don't really know how to tackle this problem either. Somehow I'm totally stuck with this one.

thx
 
cos^2x+sin^2x=1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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