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  • Integrate the functions in Exercises 1 to 24. 21.

Q: Integrate the function $\frac{2+\sin 2 x}{1+\cos 2 x} e^x$.

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Given to evaluate $\frac{2+\sin 2 x}{1+\cos 2 x} e^x$

Let $\mathrm{I}=\frac{2+\sin 2 \mathrm{x}}{1+\cos 2 \mathrm{x}} \mathrm{e}^{\mathrm{x}}$

$=\left(\frac{2+2 \sin \mathrm{x} \cos \mathrm{x}}{2 \cos ^2 \mathrm{x}}\right) \mathrm{e}^{\mathrm{x}}$

$=2 \cdot\left(\frac{1+\sin \mathrm{x} \cos \mathrm{x}}{2 \cos ^2 \mathrm{x}}\right) \mathrm{e}^{\mathrm{x}}$

$=\left(\frac{1}{\cos ^2 \mathrm{x}}+\frac{\sin \mathrm{x} \cos \mathrm{x}}{\cos ^2 \mathrm{x}}\right) \mathrm{e}^{\mathrm{x}}$

$=\left(\sec ^2 x+\tan x\right) e^x$

Now the integral becomes

$\Rightarrow \int \frac{2+\sin 2 x}{1+\cos 2 x} e^x d x=\int\left(\sec ^2 x+\tan x\right) e^x d x$

Let $\tan x=f(x)$

$\Rightarrow f^{\prime}(x)=\sec ^2 x d x$

$\Rightarrow \int \frac{2+\sin 2 x}{1+\cos 2 \mathrm{x}} \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\int\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{e}^{\mathrm{x}} \mathrm{dx}$

$=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C}$

$\Rightarrow \mathrm{I}=\mathrm{e}^{\mathrm{x}} \tan \mathrm{x}+\mathrm{C}$

Posted by

HARSH KANKARIA

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