Question

If the integral $I=\underset{0}{\overset{\pi }{\int }}\frac{{\left(\sec \right)}^{-1}\left(\sec x\right)}{1+{\left(\tan \right)}^{8}x}dx,\forall x\ne \frac{\pi }{2},$ then the value of $\left[I\right]$ is equal to (where $\left[\cdot \right]$ is the greatest integer function)

Answer 1

Answer: 2

Applying $\left(a+b-x\right)$ and adding, we get,
$\Rightarrow 2I=\pi \underset{0}{\overset{\pi }{\int }}\frac{dx}{1+{\tan }^{8}x}$
$\Rightarrow 2I=\pi \underset{0}{\overset{\pi /2}{\int }}\left(\frac{1}{1+{\tan }^{8}x}+\frac{1}{1+{\tan }^8\left(\pi -x\right)}\right)dx$
$\Rightarrow 2I=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{1+{\tan }^{8}x}=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\cos }^{8}x}{{\cos }^{8}x+{\sin }^{8}x}$
$\Rightarrow I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\cos }^{8}x}{{\cos }^{8}x+{\sin }^{8}x}$
Applying $\left(a+b-x\right)$ and adding,
$2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{8}x+{\cos }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$
$\Rightarrow 2I=\pi \underset{0}{\overset{\left(\pi \right)/2}{\int }}1dx=\pi \left(\frac{\pi }{2}\right)$
$\Rightarrow I=\frac{{\pi }^2}{4}\Rightarrow \left[I\right]=2$