TS EAMCET · Maths · Indefinite Integration
If \(\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x=f(x)+\) constant, then \(f(x)\) is equal to
- A \(e^x \cot \left(\frac{x}{2}\right)+c\)
- B \(e^{-x} \cot \left(\frac{x}{2}\right)+c\)
- C \(-e^x \cot \left(\frac{x}{2}\right)+c\)
- D \(-e^{-x} \cot \left(\frac{x}{2}\right)+c\)
Answer & Solution
Correct Answer
(C) \(-e^x \cot \left(\frac{x}{2}\right)+c\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x \\ & \quad=\int e^x\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}\right) d x \\ & \quad=\frac{1}{2} \int e^x\left(\operatorname{cosec}^2 \frac{x}{2}\right) d x-\int e^x \cot…
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