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$\int \frac{e^{x} (1 + \sin x)}{1 + \cos x} d x$ is equal to

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log⁡|tan⁡x|+C

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detailed solution

Correct option is B

∫ex(1+sin⁡x)1+cos⁡xdx=∫ex[1+2sin⁡(x/2)cos⁡(x/2)]2cos2⁡(x/2)dx=∫ex12sec2⁡x2+tan⁡x2dx=12∫exsec2⁡x2dx+∫extan⁡x2dx=12∫exsec2⁡x2dx+extan⁡x2−12∫exsec2⁡x2+C[ integrating by parts ]=extan⁡x2+C

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∫ex(1+sin⁡x)1+cos⁡xdx is equal to

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$\int \frac{e^{x} (1 + \sin x)}{1 + \cos x} d x$ is equal to