Find the value of the integral $\displaystyle \int \frac{x+\sin x}{1+\cos x} dx$ =?

 

To find the value of the integral ,consider the following trigonometric formulas:

$\displaystyle \cos 2x= 2\cos ^{2} x-1 \Rightarrow \cos x=2\cos ^{2} \frac{x}{2} -1\Rightarrow 1+\cos x=2\cos ^{2} \frac{x}{2}\\ \sin 2x= 2 \sin x \cos x\Rightarrow \sin x= 2 \sin \frac{x}{2} \cos \frac{x}{2}$ 

Hence

  

\[\begin{align*} \int \frac{x+\sin x}{1+\cos x}\; dx&= \int \left (\frac{x}{1+\cos x}\; + \frac{\sin x}{1+\cos x}\; \right ) dx\\ &=\int \left (\frac{x}{2\cos^{2} \frac{x}{2}}\; + \frac{2\sin \frac{x}{2}\cos \frac{x}2{}}{2\cos^{2} \frac{x}{2}}\; \right ) dx \\ &= \int \left (\frac{1}{2} x \sec ^{2} \frac{x}{2}\; + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\; \right ) dx \\ &= \frac{1}{2}\int x\sec ^{2} \frac{x}{2} \; dx +\int \tan \frac{x}{2} dx\\ &=\frac{1}{2}\left (x\int \sec ^{2} \frac{x}{2} \; dx -\int \left (\frac{\mathrm{d} x}{\mathrm{d} x}\int \sec ^{2}\frac{x}{2}\; dx\right ) dx \right ) +\int \tan \frac{x}{2} dx \\ &= \frac{1}{2}\left (x\; \frac{\tan \frac{x}{2}}{1/2}-\int \frac{\tan \frac{x}{2}}{1/2}dx \right )+\int \tan \frac{x}{2}dx \\&=x\tan \frac{x}{2}+C-\int \tan \frac{x}{2}dx+\int \tan \frac{x}{2}dx\\ &=x\tan \frac{x}{2}+C \end{align*}\]

\[\therefore \int \frac{(x+\sin x)}{1+\cos x} \; dx= x \; \tan \frac{x}{2} +C\]

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