1. ## Integration

Show $\int_0^\pi e^{-2t} \sin^3(t) \ dt = \frac{48}{65} \left( \dfrac{e^{2\pi} + 1}{e^{2\pi} - 1} \right) \int_0^{\pi} e^{-2t} \sin^2 (t) \ dt$ given that $\sin^3(t) = \sin(t) \left(1 - \cos^2(t)\right)$

Thanks guys

2. 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 $I_n = \int_0^\pi\!\! e^{-2t}\sin^nt\,dt$. Integrate by parts twice and you will find that if $n\geqslant 2$ then
\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 $\cos^2t = 1-\sin^2t$ for the last step). Therefore $\Bigl(1+\frac{n^2}4\Bigr)I_n = \frac{n(n-1)}4I_{n-2}$, and so $I_n = \frac{n(n-1)}{n^2+4}I_{n-2}$.

A similar calculation (integrating by parts twice) shows that $I_1 = \tfrac15(1+e^{-2\pi})$. And a straightforward integration gives $I_0 = \tfrac12(1-e^{-2\pi}).$

Therefore $I_3 = \tfrac6{13}I_1 = \tfrac6{65}(1+e^{-2\pi})$ and $I_2 = \frac14I_0 = \tfrac18(1-e^{-2\pi})$. The desired connection between $I_3$ and $I_2$ then follows.