Mathematics - Definite Integration Question with Solution | TestHub

Let, $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{\left(1+e^x\right)\left({sm}^{6}x+{\cos }^{6}x\right)} ...\left(\mathrm{i}\right)$

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

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{e^{x}dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)} ...\left(\mathrm{ii}\right)$

Adding $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$2I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{\left(1+e^x\right)dx}{\left(1+e^x\right)\left({\sin }^{6}x+{\cos }^{6}x\right)}$

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

$\underset{0}{\overset{\frac{\pi }{2}}{=\int }}\frac{dx}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-3{\sin }^{2}x{\cos }^{2}x}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{1-\frac{3}{4}{\sin }^{2}2x}$

$={\int }_0^{\frac{\pi }{2}}\frac{4{\sec }^{2}2xdx}{4\left(1+{\tan }^{2}2x\right)-3{\tan }^{2}2x}=2{\int }_0^{\frac{\pi }{4}}\frac{4{\sec }^{2}2x}{4+{\tan }^{2}2x}dx$

Let $\tan 2x=t$ and $2{\sec }^{2}2xdx=dt$

$I=4\underset{0}{\overset{\infty }{\int }}\frac{dt}{2^2+t^2}=\frac{4}{2}{\left[{\tan }^{-1}\frac{t}{2}\right]}_0^{\infty }$ $=2\left(\frac{\pi }{2}-0\right)=\pi$