The value of $\int(1-\cos x) \cdot \operatorname{cosec}^2 x \mathrm{~d} x$ is
The value of $\int(1-\cos x) \cdot \operatorname{cosec}^2 x \mathrm{~d} x$ is
- $\frac{1}{2} \tan \frac{x}{2}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\tan \frac{x}{2}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $2 \cot \frac{x}{2}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\cot \frac{x}{2}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution
Let $\mathrm{I}=\int(1-\cos x) \cdot \operatorname{cosec}^2 \mathrm{~d} x$
$=\int \frac{2 \sin ^2 \frac{x}{2}}{[\sin x]^2} \mathrm{~d} x \quad \ldots\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right]$
$\begin{aligned} &=\int \frac{2 \sin ^2 \frac{x}{2}}{\left[2 \sin \frac{x}{2} \cos \frac{x}{2}\right]^2} \mathrm{~d} x \\ & \ldots\left[\because \sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]\end{aligned}$
$\begin{aligned}
& =\frac{1}{2} \int \sec ^2 \frac{x}{2} \mathrm{~d} x \\
& =\tan \frac{x}{2}+\mathrm{c}
\end{aligned}$
Asked in: MHT CET 2023 (11 May Shift 2)
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