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