Q111 / 160 · TS EAMCET 2008 · Mathematics

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
    $e^{-x} \cot \left(\frac{x}{2}\right)+c$
  • Correct answer
    C
    $-e^x \cot \left(\frac{x}{2}\right)+c$
  • D
    $-e^{-x} \cot \left(\frac{x}{2}\right)+c$

Solution

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

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

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