MHT CET · Maths · Indefinite Integration
The value of \(\int \mathrm{e}^x\left(\frac{1-\sin x}{1-\cos x}\right) \mathrm{dx}\) is equal to
- 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)
Answer & Solution
Correct Answer
(A) \(-\mathrm{e}^x \cot \frac{x}{2}+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \mathrm{e}^x\left(\frac{1-\sin x}{1-\cos x}\right) \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^2\left(\frac{x}{2}\right)}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{1}{2} \operatorname{cosec}^2\left(\frac{x}{2}\right)-\cot \left(\frac{x}{2}\right)\right] \mathrm{d} x \\ & =-\mathrm{e}^x \cot \left(\frac{x}{2}\right)+\mathrm{c}\end{aligned}\)
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