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

$\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x=$
  1. $\mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{c}$
  2. $e^x \cot \frac{x}{2}+c$
  3. $\mathrm{e}^{\mathrm{x}} \cos \frac{\mathrm{x}}{2}+\mathrm{c}$
  4. $e^x \sin \frac{x}{2}+c$

Solution

$\begin{aligned} & I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x=\int e^x\left(\frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x}\right) d x \\ & =\int e^x\left(\frac{1}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right) d x=\int e^x\left(\frac{1}{2}\right)\left(\sec ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right) d x \\ & =\frac{1}{2} \int e^x\left[2 \tan \frac{x}{2}+\sec ^2 \frac{x}{2}\right] d x=\frac{1}{2} \cdot e^x(2)+\tan \left(\frac{x}{2}\right)+c=e^x \tan \frac{x}{2}+c\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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