MCQ
$\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$ is equal to
- A$\tan x-\cot x+C$
- ✓$-\cot x-\tan x+C$
- C$\cot x+\tan x+C$
- D$\tan x-\cot x-C$
✓
Answer
Correct option: B.
$-\cot x-\tan x+C$
$\text {Let, } I=\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$
$=\int\left(\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x \left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right]$
$=\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x$
$=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x$
$=-\cot x-\tan x+C$
$=\int\left(\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x \left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right]$
$=\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x$
$=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x$
$=-\cot x-\tan x+C$
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