A) \[{{e}^{x}}\tan \frac{x}{2}+c\]
B) \[{{e}^{x}}\cot x+c\]
C) \[\log \tan x+c\]
D) \[sin\text{ }log\text{ }x+c\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{{{e}^{x}}(1+\sin x)}{1+\cos x}}dx\] \[=\int{\frac{{{e}^{x}}(1+\sin x)}{2{{\cos }^{2}}\frac{x}{2}}}dx\] \[=\frac{1}{2}\int{\frac{{{e}^{x}}}{{{\cos }^{2}}\frac{x}{2}}}dx+\frac{1}{2}\int{\frac{{{e}^{x}}\sin x}{{{\cos }^{2}}\frac{x}{2}}}dx\] \[=\frac{1}{2}\int{{{e}^{x}}{{\sec }^{2}}\frac{x}{2}}dx+\frac{1}{2}\int{\frac{{{e}^{x}}2\left( \sin \frac{x}{2}.\cos \frac{x}{2} \right)}{{{\cos }^{2}}\frac{x}{2}}}dx\] \[=\frac{1}{2}\int{{{e}^{x}}{{\sec }^{2}}\frac{x}{2}}dx+\frac{1}{2}\int{{{e}^{x}}\tan \frac{x}{2}}dx\] \[=\frac{1}{2}\left[ \frac{{{e}^{x}}\tan \frac{x}{2}}{\frac{1}{2}}-\int{{{e}^{x}}\frac{\tan \frac{x}{2}}{\frac{1}{2}}}dx \right]+\int{{{e}^{x}}}\tan \frac{x}{2}dx\] \[={{e}^{x}}\tan \frac{x}{2}-\int{{{e}^{x}}}\tan \frac{x}{2}dx+\int{{{e}^{x}}\tan \frac{x}{2}}dx\] \[={{e}^{x}}\tan \frac{x}{2}+c\]
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