Mathematics - Definite Integration Question with Solution | TestHub

Let $I=\frac{48}{{\pi }^4}{\int }_0^{\pi }\left(\frac{3\pi }{2}{x}^2-x^3\right)\frac{\sin x}{1+{\cos }^{2}x}dx \dots \left(1\right)$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

We get $I=\frac{48}{{\pi }^4}{\int }_0^{\pi }\left(\frac{3\pi }{2}{\left(\pi -x\right)}^2-{\left(\pi -x\right)}^3\right)\frac{\sin \left(0+\mathrm{\pi }-x\right)}{1+{\cos }^2\left(0+\mathrm{\pi }-x\right)}dx$

$I=\frac{48}{{\pi }^4}{\int }_0^{\pi }\left(\frac{3{\pi }^3}{2}-3{\pi }^{2}x+\frac{3\pi x^2}{2}-{\pi }^3+x^3+3{\pi }^{2}x-3\pi x^2\right)\frac{\sin x}{1+{\cos }^{2}x}dx$ ..............equation $2$

Now adding equation (1) + equation (2)

$2I=\frac{48}{{\pi }^4}{\int }_0^{\pi }\left(\frac{3{\pi }^3}{2}-{\pi }^3\right)\frac{\sin x}{1+{\cos }^{2}x}dx$

$2I=\frac{48}{{\pi }^4}\times \frac{{\pi }^3}{2}{\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$

$2I=\frac{24}{\pi }\times 2\times {\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$

Let  $\cos x=t\Rightarrow -\sin xdx=dt$ 

So $I=\frac{24}{\pi }\times {\int }_1^0\frac{-dt}{1+t^2}$

$\Rightarrow I=\frac{-24}{\pi }{\left[{\tan }^{1}t\right]}_1^0$

$\Rightarrow I=-\frac{24}{\pi }\left[0-\frac{\pi }{4}\right]=6$