I've been integrating by parts and I would like to factor out terms on the RHS so the integral becomes the same as the LHS. \begin{align} \text{RHS} &= -\int_0^5 (5t^2 -\frac{300e^{-0.7t}}{49})(-4\pi^2 f^2 n^2 \cos(2\pi nft)) \,\mathrm{d}x \\ \text{LHS} &=\int_0^5 (e^{-0.7t})(\cos(2\pi nft)) \,\mathrm{d}x \end{align}
I've tried but have had no success as I'm not quite sure what I'm allowed to do in this situation. I've left out a large portion of the RHS, just interested in factoring this particular section.