Question

Please Datailled Explanation on how to solve
∫x^2cos(4x)dx
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Answer 1

`\displaystyle \int x^2\cos(4x)dx\hspace{5em}\textit{let's use integration by parts} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ u=x^2\implies \cfrac{du}{dx}=2x\hspace{7em}v=\displaystyle \int \cos(4x)dx\implies v=\cfrac{\sin(4x)}{4} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \int x^2\cos(4x)dx\implies \cfrac{x^2\sin(4x)}{4}-\cfrac{1}{2}\int x\sin(4x)dx\leftarrow \stackrel{ \textit{now let's again for this, use} }{\textit{integration by parts}} \\\\[-0.35em] ~\dotfill`

`~~ \hspace{5em}\displaystyle \int x\sin(4x)dx \\\\[-0.35em] ~\dotfill\\\\ u_1=x\implies \cfrac{du}{dx}=1\hspace{7em}\displaystyle v_1=\int \sin(4x)dx\implies v_1=\cfrac{-cos(4x)}{4} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \int x\sin(4x)dx\implies \cfrac{-x\cos(4x)}{4}+\cfrac{1}{4}\int \cos(4x)dx \\\\\\ \displaystyle \int x\sin(4x)dx\implies \cfrac{-x\cos(4x)}{4}+\cfrac{1}{4}\left( \cfrac{\sin(4x)}{4} \right)`

`\displaystyle \int x\sin(4x)dx\implies \cfrac{-x\cos(4x)}{4}+\cfrac{\sin(4x)}{16} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{now let's put together both the outer and nested integration by parts}}{\displaystyle \int x^2\cos(4x)dx\implies \cfrac{x^2\sin(4x)}{4}-\cfrac{1}{2}\left[ ~~ \cfrac{-x\cos(4x)}{4}+\cfrac{\sin(4x)}{16} ~~ \right]} \\\\\\ \displaystyle \int x^2\cos(4x)dx\implies \cfrac{x^2\sin(4x)}{4}+\cfrac{x\cos(4x)}{8}-\cfrac{\sin(4x)}{32}+C`