question_answer

Evaluate \[\int{\frac{2+\sin \,x}{1+\cos \,x}{{e}^{x/2}}dx.}\]

Answer:

Let  \[l=\int{\frac{2+\sin x}{1+\cos x}\,}{{e}^{x/2}}dx\] \[=\int{{{e}^{x/2}}\left( \frac{2}{1+\cos x}+\frac{\sin x}{1+\cos x} \right)\,}\,dx\] \[=\int{{{e}^{x/2}}\left( \frac{2}{2{{\cos }^{2}}\frac{x}{2}}+\frac{2\sin x\frac{x}{2}\cdot \cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}} \right)\,}\,dx\] \[=\int{{{e}^{x/2}}\left( {{\sec }^{2}}\frac{x}{2}+\tan \frac{x}{2} \right)\,}\,dx\]             Put \[\frac{x}{2}=t\Rightarrow \frac{1}{2}dx=dt\Rightarrow dx=2dt\]             \[\therefore \]      \[l=\int{{{e}^{t}}\,\,({{\sec }^{2}}t+\tan t})\,\,2dt\] \[=2\int{{{e}^{t}}\,\,({{\sec }^{2}}t+\tan t})\,\,2dt\] \[=2\int{{{e}^{t}}\,\,\{f\,(t)+f'(t)\,\,}dt\] Where,  \[f\,(t)=\tan t\] \[\therefore \]      \[I=2{{e}^{t}}f\,(t)+C=2{{e}^{t}}\tan t+C\] \[=2{{e}^{x/2}}\tan \frac{x}{2}+C\]