Question

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

Answer 1

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