Show given that
Thanks guys
I don't see any way to get directly from one of these integrals to the other. So I think that you have to evaluate each integral separately and then compare them.
Let $\displaystyle I_n = \int_0^\pi\!\! e^{-2t}\sin^nt\,dt$. Integrate by parts twice and you will find that if $\displaystyle n\geqslant 2$ then
$\displaystyle \begin{aligned}I_n &=
\Bigl[-\tfrac12e^{-2t}\sin^nt\Bigr]_0^\pi + \int_0^\pi\!\!\tfrac12 e^{-2t}n\sin^{n-1}t\cos t\,dt\\ &= \tfrac n2\Bigl[-\tfrac12e^{-2t}\sin^{n-1}t\cos t\Bigr]_0^\pi + \tfrac n4\int_0^\pi\!\! e^{-2t}\bigl((n-1)\sin^{n-2}t\cos^2 t - \sin^n t\bigr)\,dt\\ &= \tfrac n4\bigl((n-1)I_{n-2} - nI_n\bigr)\end{aligned}$
(using $\displaystyle \cos^2t = 1-\sin^2t$ for the last step). Therefore $\displaystyle \Bigl(1+\frac{n^2}4\Bigr)I_n = \frac{n(n-1)}4I_{n-2}$, and so $\displaystyle I_n = \frac{n(n-1)}{n^2+4}I_{n-2}$.
A similar calculation (integrating by parts twice) shows that $\displaystyle I_1 = \tfrac15(1+e^{-2\pi})$. And a straightforward integration gives $\displaystyle I_0 = \tfrac12(1-e^{-2\pi}).$
Therefore $\displaystyle I_3 = \tfrac6{13}I_1 = \tfrac6{65}(1+e^{-2\pi})$ and $\displaystyle I_2 = \frac14I_0 = \tfrac18(1-e^{-2\pi})$. The desired connection between $\displaystyle I_3$ and $\displaystyle I_2$ then follows.