$\displaystyle \int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}
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any help will be appreciated, thanks
$\displaystyle \frac{\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}$$\displaystyle =\frac{\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac {x}{2}+\sin^2\frac{x}{2}}=\sin\frac{x}{2}\cos\frac {x}{2}=\frac{\sin x}{2}$ so $\displaystyle \int\frac{\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}d x=-\frac{\cos x}{2}+C$