Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$$
\int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x=
$$
MathematicsIndefinite IntegrationMHT CETMHT CET 2021 (21 Sep Shift 1)
Options:
  • A $\log [\sec x(\sec x-\tan \mathrm{x})]+\mathrm{c}$
  • B $\log [\operatorname{cosec} x(\sec x+\tan x)]+c$
  • C $\log [\sec x(\sec x+\tan x)]+c$
  • D $\log [\sec \mathrm{x}+\tan \mathrm{x}]+\mathrm{c}$
Solution:
2837 Upvotes Verified Answer
The correct answer is: $\log [\sec x(\sec x+\tan x)]+c$
Let $\left.I=\int[1+2 \tan x(\tan x+\sec x)]^{1 / 2} d x=\int 1+2 \tan ^2 x+2 \tan x \sec x\right)^{1 / 2} d x$
$$
\int\left[\left(1+\tan ^2 x\right)+\tan ^2 x+2 \sec x \tan x\right]^{1 / 2} d x=\int\left(\sec ^2 x+\tan ^2 x+2 \sec x \tan x\right)^{1 / 2} d x
$$

$$
\begin{aligned}
& =\int\left[(\sec x+\tan x)^2\right]^{1 / 2} d x=\int(\sec x+\tan x) d x=\int \sec x d x+\int \tan x d x \\
& =\log |\sec x+\tan x|-\log |\cos x|+c \quad=\log \frac{|\sec x+\tan x|}{|\cos x|}+c \\
& =\log [\sec x(\sec x+\tan x)]+c
\end{aligned}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.