Q85 / 160 · AP EAMCET 2010 · Mathematics
If $\int(1-\cos x) \operatorname{cosec}^2 x d x=f(x)+c$, then $f(x)$ is equal to
- Correct answerA$\tan \frac{x}{2}$
- B$\cot \frac{x}{2}$
- C$2 \tan \frac{x}{2}$
- D$\frac{1}{2} \tan \frac{x}{2}$
Solution
$\int(1-\cos x) \operatorname{cosec}^2 x d x$
$=\int\left(2 \sin ^2 \frac{x}{2}\right) \cdot \frac{1}{\sin ^2 x} d x$
$\left[\because \cos x=1-2 \sin ^2 \frac{x}{2}\right]$
$=2 \int \frac{\sin ^2 \frac{x}{2}}{\left(2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\right)^2} d x$
$\left[\because \sin x=2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\right]$
$=2 \int \frac{\sin \frac{x}{2} \cdot \sin \frac{x}{2}}{2 \cdot 2 \sin ^2 \frac{x}{2} \cdot \cos ^2 \frac{x}{2}} d x$
$=\frac{1}{2} \int \sec ^2 \frac{x}{2} d x=\frac{1}{2} \tan \frac{x}{2} \cdot 2+c$
$=\tan \frac{x}{2}+c=f(x)+c$
$\Rightarrow \quad f(x)=\tan \frac{x}{2}$
Hence, the correct answer is option: 'A'.