The value of $\int \mathrm{e}^x\left(\frac{1-\sin x}{1-\cos x}\right) \mathrm{dx}$ is equal to

The value of $\int \mathrm{e}^x\left(\frac{1-\sin x}{1-\cos x}\right) \mathrm{dx}$ is equal to
  1. $-\mathrm{e}^x \cot \frac{x}{2}+\mathrm{c}$, (where c is a constant of integration)
  2. $\mathrm{e}^x \cot \frac{x}{2}+\mathrm{c}$, (where c is a constant of integration)
  3. $\mathrm{e}^x \operatorname{cosec} \frac{x}{2}+\mathrm{c}$, ( where c is a constant of integration)
  4. $-\mathrm{e}^x \operatorname{cosec} \frac{x}{2}+\mathrm{c}$, (where c is a constant of integration)

Solution

$\begin{aligned} & \int \mathrm{e}^x\left(\frac{1-\sin x}{1-\cos x}\right) \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^2\left(\frac{x}{2}\right)}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{1}{2} \operatorname{cosec}^2\left(\frac{x}{2}\right)-\cot \left(\frac{x}{2}\right)\right] \mathrm{d} x \\ & =-\mathrm{e}^x \cot \left(\frac{x}{2}\right)+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 2)

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