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Question

Question: \[\int_{}^{}{\frac{x + \sin x}{1 + \cos x}dx =}\]...

∫x+sin⁡x1+cos⁡xdx=\int_{}^{}{\frac{x + \sin x}{1 + \cos x}dx =}

A

−xtan⁡x/2+c- x\tan x/2 + c

B

xtan⁡x/2+cx\tan x/2 + c

C

xtan⁡x+cx\tan x + c

D

12xtan⁡x+c\frac{1}{2}x\tan x + c

Answer

xtan⁡x/2+cx\tan x/2 + c

Explanation

Solution

∫x+sin⁡x1+cos⁡xdx=12∫xsec⁡2x2dx+∫tan⁡x2dx=12xtan⁡x/21/2−∫tan⁡x2dx+∫tan⁡x2dx=xtan⁡x2+c\int_{}^{}\frac{x + \sin x}{1 + \cos x}dx = \frac{1}{2}\int_{}^{}{x\sec^{2}\frac{x}{2}dx +}\int_{}^{}{\tan\frac{x}{2}dx} = \frac{1}{2}\frac{x\tan x/2}{1/2} - \int_{}^{}{\tan\frac{x}{2}dx + \int_{}^{}{\tan\frac{x}{2}dx = x\tan\frac{x}{2} + c}}

Trick : By inspection, ddx{xtan⁡x2+c}\frac{d}{dx}\left\{ x\tan\frac{x}{2} + c \right\}

=xsec⁡2x2+tan⁡x2=12[xcos⁡2x/2+2sin⁡x/2cos⁡x/2]=x+sin⁡x1+cos⁡x= x\sec^{2}\frac{x}{2} + \tan\frac{x}{2} = \frac{1}{2}\left\lbrack \frac{x}{\cos^{2}x/2} + \frac{2\sin x/2}{\cos x/2} \right\rbrack = \frac{x + \sin x}{1 + \cos x}

Hence the result.