\(\int e^{x / 2}\left(\frac{2+\sin x}{1+\cos x}\right) d x=\)

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

Solution

\(\begin{aligned} & \int e^{\frac{x}{2}}\left(\frac{2+\sin x}{1+\cos x}\right) d x \\ & \int e^{\frac{x}{2}}\left(\frac{2+\frac{2 \tan x / 2}{1+\tan ^2 x / 2}}{1+\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}}\right) d x \\ & \int e^{\frac{x}{2}}\left(\frac{2+2 \tan ^2 x / 2+2 \tan x / 2}{1+\tan ^2 x / 2+1-\tan ^2 x / 2}\right) d x \end{aligned}\) \(\begin{gathered} \int e^{\frac{x}{2}} \cdot 2\left(\frac{1+\tan ^2 x / 2+\tan x / 2}{2}\right) d x \\ \int e^{\frac{x}{2}}\left(\sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right) d x \\ \text {Put, } \frac{x}{2}=t \\ d x=2 d t \\ \int e^t\left(\sec ^2 t+\tan t\right) \cdot 20 d t \\ 2 \cdot e^t \tan t+c \\ 2 \cdot e^{\frac{x}{2}} \tan \frac{x}{2}+c \end{gathered}\) Hence, option (b) is correct.

Asked in: AP EAMCET 2020 (18 Sep Shift 2)

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