MHT CET · Maths · Indefinite Integration
\(\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x} d x=\)
- A \(\log \left|\tan ^{2} x\right|+c\)
- B \(\log \left|\sec ^{2} x\right|+c\)
- C \(\log |\tan x|+c\)
- D \(\log |\sec x|+c\)
Answer & Solution
Correct Answer
(A) \(\log \left|\tan ^{2} x\right|+c\)
Step-by-step Solution
Detailed explanation
\(I =\int \frac{\sin 2 x}{\sin ^{2} x \cos ^{2} x}=\int \frac{2 \sin x \cos x d x}{\sin ^{2} x \cos ^{2} x}=2 \int \frac{1}{\sin x \cos x} d x=2 \int \frac{2}{2 \sin x \cos x} d x \)
\( =2 \times 2 \int \frac{1}{\sin 2 x} d x=4 \int \operatorname{cosec} 2 x d x \)
\( =2 \log |\tan x|+c=\log \left|\tan ^{2} x\right|+c\)
\( =2 \times 2 \int \frac{1}{\sin 2 x} d x=4 \int \operatorname{cosec} 2 x d x \)
\( =2 \log |\tan x|+c=\log \left|\tan ^{2} x\right|+c\)
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