\(\int {\frac{{{e^x}\left( {1 + \sin x} \right)}}{{1 + \cos x}}} dx\) समान है: 

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UP TGT Mathematics 2016 Official Paper
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  1. \(\log \tan x + c\)
  2. \({e^x}\tan \frac{x}{2} + c\)
  3. \(\sin \log x + c\)
  4. \({e^x} \cot x + c\)

Answer (Detailed Solution Below)

Option 2 : \({e^x}\tan \frac{x}{2} + c\)
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Detailed Solution

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सूत्र:

\(\int {e}^x [f(x) + f'(x)] dx = e^x f(x) + C\)

गणना:

माना  \(I = \int e^x . \frac{1 + sinx}{1 + cos x} dx\)

⇒ \(I = \int e^x . \frac{1 + 2sin\frac{x}{2}cos \frac{x}{2}}{ 2cos^2 \frac{x}{2}} dx\)

⇒ \(I = \int e^x . [\frac{1}{2 cos^2\frac{x}{2}} + \frac{2 sin\frac{x}{2} cos\frac{x}{2}}{2 cos^2\frac{x}{2}}] dx\)

⇒ \(I = \int e^x . [\frac{1}{2} sec^2 \frac{x}{2} + tan \frac{x}{2}] dx\)

उपर्युक्त समाकल \(\int {e}^x [f(x) + f'(x)] dx\) रूप का है 

\(f(x) = tan \frac{x}{2}\)

 \(f'(x) = \frac{1}{2} sec^2 \frac{x}{2}\)

 \(∴\ I = e^x f(x) + c\)

\(I = e^x tan \frac{x}{2} + c\)

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