A) \[2.{{e}^{x/2}}\,\tan \frac{x}{2}+C\]
B) \[{{e}^{x/2}}\,\tan \,x+C\]
C) \[\frac{1}{2}{{e}^{x/2}}\,\sin x+C\]
D) \[\frac{1}{2}\,\,{{e}^{x/2}}\,\sin \frac{x}{2}+C\]
Correct Answer: A
Solution :
Let \[l=\int{\frac{2+\sin x}{1+\cos x}}.\,\,{{e}^{x/2}}\,dx\] \[\Rightarrow \] \[l=\int{\frac{2+\frac{2\,\tan \,x/2}{1+{{\tan }^{2}}x/2}}{1+\frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2}}}\,.\,\,{{e}^{-x/2}}dx\] \[\Rightarrow \] \[l=\frac{2{{\tan }^{2}}\frac{x}{2}+2+2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}-{{\tan }^{2}}\frac{x}{2}+1}\,\,.\,{{e}^{x/2}}\,dx\] \[\Rightarrow \] \[l=2\int{\frac{{{\tan }^{2}}\frac{x}{2}+\tan \frac{x}{2}+1}{2}}.{{e}^{x/2}}\,\,dx\] \[\Rightarrow \] \[l=\int{{{\tan }^{2}}\frac{x}{2}.{{e}^{x/2}}dx+\int{\tan \frac{x}{2}.{{e}^{x/2}}\,dx}}\] \[+\int{{{e}^{x/2}}\,\,dx}\] \[\Rightarrow \] \[l=\int{\underset{II}{\mathop{{{\sec }^{2}}}}\,}\,x/2.{{e}^{x/2}}\,dx\] \[-\int{{{e}^{x/2}}\,dx+\int{\tan \frac{x}{2}.{{e}^{x/2}}\,dx+\int{{{e}^{x/2}}\,dx}}}\] \[\Rightarrow \] \[l=2{{e}^{x/2}}.\tan \frac{x}{2}-\int{\frac{1}{2}{{e}^{x/2}}.\tan \frac{x}{2}.2dx}\] \[+\int{\tan \,\frac{x}{2}.\,{{e}^{x/2}}\,dx+C}\] \[\Rightarrow \] \[l=2{{e}^{x/2}}.\tan \frac{x}{2}-\int{{{e}^{x/2}}.\tan \frac{x}{2}\,dx}\] \[+\int{{{e}^{x/2}}.\tan \frac{x}{2}dx+C}\] \[\Rightarrow \] \[l=2{{e}^{x/2}}.\tan \frac{x}{2}+C\]
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