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Question

$$\displaystyle I= \int e^{x}\frac{\left ( 2+\sin 2x \right )}{\left ( 1+\cos 2x \right )}dx.$$


A
exsinx.
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B
excosx.
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C
extanx.
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D
excos2x.
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Solution

The correct option is B $$\displaystyle e^{x}\tan x.$$
$$\displaystyle I= \int e^{x}\frac{\left ( 2+\sin 2x \right )}{\left ( 1+\cos 2x \right )}dx.$$
$$\displaystyle = \int \left [ \frac{e^{x}.2}{2\cos ^{2}x}+e^{x}\frac{2\sin x\cos x}{2\cos^{2} x} \right ]dx$$
$$\displaystyle \int e^{x}\sec^{2}xdx+\int e^{x}\tan xdx.$$
Integrate first by parts
$$\displaystyle I=e^{x}\tan x-\int e^{x}\tan xdx+\int e^{x}\tan xdx.$$
$$\displaystyle = e^{x}\tan x.$$ The last two integrals cancel.
Above is of the form
$$\displaystyle \int e^{x}\left [ f\left ( x \right )+f'\left ( x \right )\right ]dx=e^{x}f\left ( x \right ),$$
where $$\displaystyle f\left ( x \right )=\tan x$$ and $$\displaystyle f\left ( x \right )=\sec^{2}x.$$

Mathematics

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