The value of $\int\limits_0^{\pi / 4}\left(\tan ^n x+\tan ^{n-2} x\right) d\left(x-\frac{[x]}{1 !}+\frac{[x]^2}{2 !}-\frac{[x]^3}{3 !}+... \right)$ where [x] is greatest integer function, is
$[x \in[0, \pi / 4]]$
Answer & explanation
Correct answer: option 3
For $x \in[0, \pi / 4],[x]=0$
∴ $\int\limits_0^{\pi / 4}\left(\tan ^n x+\tan ^{n-2} x\right) d\left(x-\frac{[x]}{1 !}+\frac{[x]^2}{2 !}-\frac{[x]^3}{3 !}+...\right)$
$=\int\limits_0^{\pi / 4} \tan ^{n-2} x \sec ^2 x d x=\left[\frac{\tan ^{n-1} x}{n-1}\right]_0^{\pi / 4}=\frac{1}{n-1}$