Q99 / 180 · KCET 2015 · Mathematics

\( \int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x \) is

  • Correct answer
    A
    \( e^{x} \tan \left(\frac{x}{2}\right)+C \)
  • B
    \( \tan \left(\frac{x}{2}\right)+C \)
  • C
    \( e^{x}+C \)
  • D
    \( e^{x} \sin x+C \)

Solution

Given that,\( I=\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x \) \( =\int e^{x}\left(\frac{1+2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{2 \cos ^{2}\left(\frac{x}{2}\right)}\right) d x \) \( =\int e^{x}\left[\frac{1}{2 \cos ^{2}\left(\frac{x}{2}\right)}+\frac{\sin \left(\frac{x}{2}\right)}{\cos ^{2}\left(\frac{x}{2}\right)}\right] d x \) \( =\int e^{x}\left(\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right)+\tan \left(\frac{x}{2}\right)\right) d x \) Let \( u=\tan \left(\frac{x}{2}\right) \operatorname{then} \frac{d u}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \) So, \( I=\int e^{x}\left[u+\frac{d u}{d x}\right] d x=e^{x} u \) \( =e^{x} \tan \left(\frac{x}{2}\right)+c \)

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

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