Mathematics - Indefinite Integration Question with Solution | TestHub

$I_n={\int }_0^{\pi /4}({\tan }^n(x-[x])+{\tan }^{n-2}(x-[x]))dx={\int }_0^{\pi /4}({\tan }^{n}x+{\tan }^{n-2}x)dx$

$$\{\because 0\le x\le \frac{\pi }{4}\Rightarrow [x]=0\}$$

$={\int }_0^{\pi /4}{\tan }^{n-2}x({\tan }^{2}x+1)dx={\int }_0^{\pi /4}{\sec }^{2}x\cdot {\tan }^{n-2}xdx$

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

$I_n={\left(\frac{t^{n-1}}{n-1}\right)}_0^1\Rightarrow \frac{1}{n-1}$.

$I_2,I_3,I_4,\dots =\frac{1}{1},\frac{1}{2},\frac{1}{3},\dots$.

Therefore, $I_2,I_3,I_4,\dots$ are in H.P.

$I_{100}=\frac{1}{99}$, $I_{99}=\frac{1}{98}$.

$\frac{1}{I_{100}}-\frac{1}{I_{99}}=99-98=1$.

${\sum }_{i=2}^n\frac{1}{I_i}=\frac{1}{I_2}+\frac{1}{I_3}+\dots +\frac{1}{I_n}=1+2+3+\dots +(n-1)=\frac{n(n-1)}{2}$.

Hence, options (A), (B) & (D) are correct.