Q113 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Mathematics

$\int \frac{1 + \tan x \tan (x + a)}{\tan x \tan (x + a)} d x =$

A
$\tan a (\log (\sec (x + a)) + logsec x + C)$
Correct answer
B
$\cot a (\log |\sin x| - \log |\sin (x + a)|) + C$
C
$\tan a (\log (\frac{\cos x}{\sin (x + a)})) + C$
D
$\cot a (\log \frac{\sin (x + a)}{\cos (x + a)}) + C$

Solution

$I = \int \frac{1 + \tan x \tan (x + a)}{\tan x \tan (x + a)} d x$
We know $\tan a = \tan [(x + a) - x] = \frac{\tan (x + a) - \tan x}{1 + \tan x \tan (x + a)}$
So, $I = \cot a \int \frac{\tan (x + a) - \tan x}{\tan (x + a) \tan x} d x$
$= \cot a \int [\frac{1}{\tan x} - \frac{1}{\tan (x + a)}] d x$
$= \cot a (\log |\sin x| - \log |\sin (x + a)|) + C$

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

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