Read the following passage carefully:If f(x) is a differentiable function, then ∫exf(x)+f′(x)dx=exf(x)+c Using the above information. Simplify the following.If x∈0,π2, then ∫e−x/2⋅1−Sin⁡x1+Cos⁡xdx=

# Read the following passage carefully:If f(x) is a differentiable function, then $\int {\mathrm{e}}^{\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\right]\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{c}$ Using the above information. Simplify the following.If

1. A

${\mathrm{e}}^{-\mathrm{x}/2}\mathrm{Sec}x+\mathrm{c}$

2. B

$-{\mathrm{e}}^{-\mathrm{x}/2}\mathrm{Sec}\frac{\mathrm{x}}{2}+\mathrm{c}$

3. C

${\mathrm{e}}^{\mathrm{x}/2}\cdot \mathrm{Sec}\frac{\mathrm{x}}{2}+\mathrm{c}$

4. D

$-{\mathrm{e}}^{\mathrm{x}/2}\cdot \mathrm{Sec}\frac{\mathrm{x}}{2}+\mathrm{c}$

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### Solution:

$\begin{array}{l}\mathrm{G}.\mathrm{I}=±\int {\mathrm{e}}^{-\mathrm{x}/2}\left[\frac{\mathrm{sin}\frac{\mathrm{x}}{2}-\mathrm{cos}\frac{\mathrm{x}}{2}}{2{\mathrm{cos}}^{2}\frac{\mathrm{x}}{2}}\right]\mathrm{dx}\\ =±\int {\mathrm{e}}^{-\frac{\mathrm{x}}{2}}\left[\left(-\frac{1}{2}\right)\left(\mathrm{sec}\frac{\mathrm{x}}{2}\right)+\left(\frac{1}{2}\mathrm{sec}\frac{\mathrm{x}}{2}\mathrm{tan}\frac{\mathrm{x}}{2}\right)\right]\mathrm{dx}\\ =±{\mathrm{e}}^{-\frac{\mathrm{x}}{2}}\left(\mathrm{sec}\frac{\mathrm{x}}{2}\right)\end{array}$

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