$\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x} d x=$

$\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x} d x=$
  1. $\log \left|\tan ^{2} x\right|+c$
  2. $\log \left|\sec ^{2} x\right|+c$
  3. $\log |\tan x|+c$
  4. $\log |\sec x|+c$

Solution

$\begin{aligned} I &=\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x}=\int \frac{2 \sin x \cos x d x}{\sin ^{2} x \cos ^{2} x}=2 \int \frac{1}{\sin x \cos x} d x=2 \int \frac{2}{2 \sin x \cos x} d x \\ &=2 \times 2 \int \frac{1}{\sin 2 x} d x=4 \int \operatorname{cosec} 2 x d x \\ &=2 \log |\tan x|+c=\log \left|\tan ^{2} x\right|+c \end{aligned}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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