Question

${\int }_0^{\frac{\pi }{2}}\frac{x\sin 2xdx}{{\cos }^{4}x+{\sin }^{4}x}$

• A. $\frac{{\pi }^2}{8}$
• B. $\frac{3{\pi }^2}{8}$
• C. $\frac{{\pi }^2}{4}$
• D. $\frac{3{\pi }^2}{4}$

Answer 1

The correct option is A $\frac{{\pi }^2}{8}$

Let $I={\int }_0^{\frac{\pi }{2}}\frac{x\sin 2xdx}{{\cos }^{4}x+{\sin }^{4}x}={\int }_0^{\frac{\pi }{2}}\frac{2x\sin x\cos xdx}{{\cos }^{4}x+{\sin }^{4}x}$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{2(\frac{\pi }{2}-x)\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)dx}{{\cos }^4(\frac{\pi }{2}-x)+{\sin }^4(\frac{\pi }{2}-x)}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{2(\frac{\pi }{2}-x)\sin x\cos xdx}{{\cos }^{4}x+{\sin }^{4}x}=\pi {\int }_0^{\frac{\pi }{2}}\frac{\sin x\cos xdx}{{\cos }^{4}x+{\sin }^{4}x}-{\int }_0^{\frac{\pi }{2}}\frac{2x\sin x\cos xdx}{{\cos }^{4}x+{\sin }^{4}x}=\pi {\int }_0^{\frac{\pi }{2}}\frac{\sin x\cos xdx}{{\cos }^{4}x+{\sin }^{4}x}-I\Rightarrow 2I=\pi {\int }_0^{\frac{\pi }{2}}\frac{\sin x\cos xdx}{{\cos }^{4}x+{\sin }^{4}x}=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\left({\tan }^{2}x\right)}^2}d\left({\tan }^{2}x\right)\Rightarrow 2I=\frac{\pi }{2}{\left[{\tan }^{-1}t\right]}_0^{\infty }=\frac{\pi }{2}({\tan }^{-1}\infty -{\tan }^{-1}0)\Rightarrow I=\frac{{\pi }^2}{8}$