Question

If $\int \frac{\sec ^2 x}{\tan ^{2017} x+\tan x} d x=g(x)+c$, where $g\left(\frac{\pi}{4}\right)=\frac{-\ln 2}{2016}$, then $\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text{Lim}} g(x)$ 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, $I=\int \frac{\sec ^2 x}{\tan ^{2017} x+\tan x} d x$
Let $ \tan x = t \Rightarrow \sec ^2 xdx = dt$
$I=\int \frac{1}{t^{2017}+t} d t=\int \frac{\frac{1}{t^{2017}}}{1+\frac{1}{t^{2016}}} d t$
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 \rightarrow \frac{-}{2}^{-}} {\text{Lim}} g ( x )=0 $