Mathematics - Definite Integration Question with Solution | TestHub

Let, $I={\int }_0^{\mathrm{\pi }}\frac{x^2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

Now using the property ${\int }_0^{2a}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(2a-x\right)dx$ we get,

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

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

$\Rightarrow I=2\pi \underset{I_1}{\underbrace{{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{x·\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx}}-{\mathrm{\pi }}^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

Now, solving $I_1={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{x·\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx ....\left(1\right)$

$\Rightarrow I_1={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\frac{\mathrm{\pi }}{2}-x\right)·\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx ....\left(2\right)$

Adding both equations we get,

$\Rightarrow 2I_1=\frac{\mathrm{\pi }}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$\Rightarrow I_1=\frac{\mathrm{\pi }}{4}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

Now, putting the value in $I$ we get,

$\Rightarrow I=2\pi ·\frac{\mathrm{\pi }}{4}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{x·\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx-{\mathrm{\pi }}^2{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{1-2{\sin }^{2}x{\cos }^{2}x}dx$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\displaystyle \frac{1}{2}}\sin 2x}{1-{\displaystyle \frac{1}{2}}{\sin }^{2}2x}dx$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\displaystyle \sin 2x}}{2-{\sin }^{2}2x}dx$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\displaystyle \sin 2x}}{1+{\cos }^{2}2x}dx$

Now, let $\cos 2x=t\Rightarrow -2\sin 2xdx=dt$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{2}{\int }_1^{-1}\frac{{\displaystyle -\frac{{\displaystyle 1}}{2}}}{1+t^2}dt$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{4}{\int }_{-1}^1\frac{{\displaystyle 1}}{1+t^2}dt$

$\Rightarrow I=-\frac{{\mathrm{\pi }}^2}{4}·\frac{\mathrm{\pi }}{2}=-\frac{{\mathrm{\pi }}^3}{8}$

Hence, $\frac{120}{{\mathrm{\pi }}^3}\left|{\int }_0^{\mathrm{\pi }}\frac{x^2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\right|=\frac{120}{{\mathrm{\pi }}^3}\times \frac{{\mathrm{\pi }}^3}{8}=15$