Q112 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

$\int \frac{2 \cos 2 x}{(1+\sin 2 x)(1+\cos 2 x)} d x=$

  • A
    $2 \tan x+\log (1+\tan x)+c$
  • B
    $\tan x-2 \log (1+\tan x)+c$
  • C
    $2 \log (1+\tan x)+\tan x+c$
  • Correct answer
    D
    $2 \log (1+\tan x)-\tan x+c$

Solution

$I=\int \frac{2 \cos 2 x}{(1+\sin 2 x)(1+\cos 2 x)} d x$ $\begin{aligned} & =2 \int \frac{\frac{1-\tan ^2 x}{1+\tan ^2 x}}{\left(1+\frac{2 \tan x}{1+\tan ^2 x}\right)\left(1+\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)} d x \\ & =\int \frac{(1-\tan x) \sec ^2 x}{(1+\tan x)} d x \cdot \text { Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \\ & =\int\left(\frac{1-t}{1+t}\right) d t=\int\left(-1+\frac{2}{1+t}\right) d t \\ & =-\tan x+2 \log (1+\tan x)+c .\end{aligned}$

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

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