Q93 / 150 · MHT CET 2024 (03 May Shift 2) · Mathematics

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

  • Correct answer
    A
    $-\mathrm{e}^x \cot \frac{x}{2}+\mathrm{c}$, (where c is a constant of integration)
  • B
    $\mathrm{e}^x \cot \frac{x}{2}+\mathrm{c}$, (where c is a constant of integration)
  • C
    $\mathrm{e}^x \operatorname{cosec} \frac{x}{2}+\mathrm{c}$, ( where c is a constant of integration)
  • D
    $-\mathrm{e}^x \operatorname{cosec} \frac{x}{2}+\mathrm{c}$, (where c is a constant of integration)

Solution

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

Hence, the correct answer is option: 'A'.

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