Question

How do you find the integral of $\left[ {{x^4}\sin x\,dx} \right]$?

Answer 1

Hint:In this question, we have to do by-parts integration. But here we have to do the by-parts integration not once but several times till we get to the final answer. So, for that we should know the formula of by-parts integration and how we apply it.

Formula used: $\int {u\,v\,dx = u\int {v\,dx - \int {\left( {\frac{{du}}{{dx}}\int {vdx} } \right)dx} } } $

Complete step by step answer:
In the above question, it is given that we have to find the integral of $\left[ {{x^4}\sin x\,dx} \right]$.
In this question, we need to integrate by parts repeatedly, to reduce the degree of x.
$\int {{x^4}\sin x\,dx = } {x^4}\int {\sin xdx - \int {\left[ {\frac{d}{{dx}}{x^4}\int {\sin xdx} } \right]} } $
$ \Rightarrow {x^4}\left( { - \cos x} \right) - \left[ {\int {4{x^3}\left( { - \cos x} \right)dx} } \right]$
Now, on simplification we get
$ \Rightarrow - {x^4}\cos x + \left[ {\int {4{x^3}\left( {\cos x} \right)dx} } \right]$
$ \Rightarrow - {x^4}\cos x + 4{x^3}\int {\cos dx - \left[ {\int {\frac{d}{{dx}}\left( {4{x^3}} \right)\int {\cos xdx} } } \right]} $
$ \Rightarrow - {x^4}\cos x + 4{x^3}\sin x - \int {12{x^2}\sin xdx} $
$ \Rightarrow - {x^4}\cos x + 4{x^3}\sin x + 12{x^2}\cos x - \int {24x\cos xdx} $
$ \Rightarrow - {x^4}\cos x + 4{x^3}\sin x + 12{x^2}\cos x - 24x\sin x + 24\int {\sin xdx} $
$ \Rightarrow - {x^4}\cos x + 4{x^3}\sin x + 12{x^2}\cos x - 24x\sin x + 24\left( { - \cos x} \right) + C$
$ \Rightarrow - {x^4}\cos x + 4{x^3}\sin x + 12{x^2}\cos x - 24x\sin x - 24\cos x + C$
Therefore, the value of the required integral is $ \Rightarrow - {x^4}\cos x + 4{x^3}\sin x + 12{x^2}\cos x - 24x\sin x - 24\cos x + C$.

Note:Integration by parts is a special technique of integration of two functions when they are multiplied. This method is also termed as partial integration. Another method to integrate a given function is integration by substitution method. These methods are used to make complicated integrations easy.