Question

If $\int \frac{{\sec }^{2}x}{{\tan }^{2017}x+\tan x}dx=g(x)+c$, where $g\left(\frac{\pi }{4}\right)=\frac{-\ln 2}{2016}$, then $\underset{x\to \frac{{\pi }^{-}}{2}}{\mathrm{Lim}}g(x)$ is equal to (where $c$ is constant of integration)

• A. 0
• B. 1
• C. $\frac{-1}{2016}$
• D. $-\ln 2$

Answer 1

Answer: (A)

Given integral, $I=\int \frac{{\sec }^{2}x}{{\tan }^{2017}x+\tan x}dx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\int \frac{1}{t^{2017}+t}dt=\int \frac{\frac{1}{t^{2017}}}{1+\frac{1}{t^{2016}}}dt$
Let $1+\frac{1}{t^{2016}}=z\Rightarrow \frac{-2016}{t^{2017}}dt=dz$
$\therefore I=\frac{-1}{2016}\ln \left(1+\frac{1}{t^{2016}}\right)+c$
$\therefore I=\frac{-1}{2016}\ln \left(1+\frac{1}{{\tan }^{2016}x}\right)+c$
$\therefore \underset{x\to {\frac{-}{2}}^{-}}{\mathrm{Lim}}g(x)=0$