Mixed Integral between Trig and Exp asked in the math group

Exercise:

Integrate , $\int{{{e}^{x}}\frac{1+\sin x}{1+\cos x}\,dx}$

Solution: we know that $1+\cos 2\theta =2{{\cos }^{2}}\theta $ so put $\theta =\frac{x}{2}$ to get $1+\cos x=2{{\cos }^{2}}\left( \frac{x}{2} \right)$

Also we have $\sin 2\theta =2\sin \theta \cos \theta $ so by taking $\theta =\frac{x}{2}$ we get $\sin x=2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)$

Hence $\int{{{e}^{x}}\frac{1+\sin x}{1+\cos x}dx}=\int{{{e}^{x}}\frac{1+2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)}{2{{\cos }^{2}}\left( \frac{x}{2} \right)}\,dx}=\int{\frac{1}{2}{{e}^{x}}{{\sec }^{2}}\left( \frac{x}{2} \right)\,dx}+\int{{{e}^{x}}\tan \left( \frac{x}{2} \right)\,dx}$

Let $u=\tan \left( \frac{x}{2} \right)\,\,\And \,\,dv={{e}^{x}}\Leftrightarrow du=\frac{1}{2}{{\sec }^{2}}\left( \frac{x}{2} \right)dx\,\,\And \,v={{e}^{x}}$

So $\int{{{e}^{x}}\tan \left( \frac{x}{2} \right)\,dx}={{e}^{x}}\tan \left( \frac{x}{2} \right)-\int{\frac{1}{2}{{e}^{x}}{{\sec }^{2}}\left( \frac{x}{2} \right)\,dx}$

Hence $\int{{{e}^{x}}\frac{1+\sin x}{1+\cos x}\,dx}=\frac{1}{2}\int{{{e}^{x}}{{\sec }^{2}}\left( \frac{x}{2} \right)\,dx}+{{e}^{x}}\tan \left( \frac{x}{2} \right)-\frac{1}{2}\int{{{e}^{x}}{{\sec }^{2}}\left( \frac{x}{2} \right)\,dx}$

Thus $\int{{{e}^{x}}\frac{1+\sin x}{1-\cos x}\,dx}={{e}^{x}}\tan \left( \frac{x}{2} \right)\,+c$