Q76 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

$$ \int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x= $$

  • A
    $\log [\sec x(\sec x-\tan \mathrm{x})]+\mathrm{c}$
  • B
    $\log [\operatorname{cosec} x(\sec x+\tan x)]+c$
  • Correct answer
    C
    $\log [\sec x(\sec x+\tan x)]+c$
  • D
    $\log [\sec \mathrm{x}+\tan \mathrm{x}]+\mathrm{c}$

Solution

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

Hence, the correct answer is option: 'C'.

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