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Question: \(\int_{}^{}{\frac{e^{x}\left( 1 + \sin x \right)}{1 + \cos x}dx}\)is equal to...

∫ex(1+sin⁡x)1+cos⁡xdx\int_{}^{}{\frac{e^{x}\left( 1 + \sin x \right)}{1 + \cos x}dx}is equal to

A

log⁡∣tan⁡x∣+C\log|\tan x| + C

B

extan⁡(x2)+Ce^{x}\tan\left( \frac{x}{2} \right) + C

C

sin⁡excot⁡x+C\sin e^{x}\cot x + C

D

excot⁡x+Ce^{x}\cot x + C

Answer

extan⁡(x2)+Ce^{x}\tan\left( \frac{x}{2} \right) + C

Explanation

Solution

∫ex(1+sin⁡x)1+cos⁡xdx\int_{}^{}{\frac{e^{x}\left( 1 + \sin x \right)}{1 + \cos x}dx}= ∫ex[1+2sin⁡(x2)cos⁡(x2)]2cos⁡2(x2)dx\int_{}^{}{\frac{e^{x}\left\lbrack 1 + 2\sin\left( \frac{x}{2} \right)\cos\left( \frac{x}{2} \right) \right\rbrack}{2\cos^{2}\left( \frac{x}{2} \right)}dx}

= ∫ex(12sec⁡2x2+tan⁡x2)dx=12∫exsec⁡2x2dx+∫extan⁡x2dx\int_{}^{}{e^{x}\left( \frac{1}{2}\sec^{2}\frac{x}{2} + \tan\frac{x}{2} \right)dx =}\frac{1}{2}\int_{}^{}{e^{x}\sec^{2}\frac{x}{2}dx + \int_{}^{}{e^{x}\tan\frac{x}{2}dx}}

= 12∫exsec⁡2x2+extan⁡x2−12∫exsec⁡2x2+C\frac{1}{2}\int_{}^{}{e^{x}\sec^{2}\frac{x}{2} + e^{x}\tan\frac{x}{2} - \frac{1}{2}\int_{}^{}{e^{x}\sec^{2}\frac{x}{2} + C}}

[integrating by parts]

=extan⁡x2+C.= e^{x}\tan\frac{x}{2} + C.