A) \[{{\log }_{e}}\,|\tan x|+K\]
B) \[{{e}^{x}}\,\tan \,\left( \frac{x}{2} \right)+K\]
C) \[{{e}^{x}}\,\tan \,x+K\]
D) \[{{e}^{x}}\,{{\log }_{e}}\,|\sec \,x|+K\]
Correct Answer: B
Solution :
Let \[I=\int{\frac{{{e}^{x}}(1+\sin x)}{1+\cos \,x}}\,dx\] \[=\int{\frac{{{e}^{x}}\left( 1+2\sin \frac{x}{2}\cos \frac{x}{2} \right)}{2{{\cos }^{2}}\frac{x}{2}}}\,dx\] \[=\int{\frac{1}{2}\,{{e}^{x}}\,{{\sec }^{2}}\,\frac{x}{2}dx+\int{\underset{I}{\mathop{{{e}^{x}}}}\,\underset{II}{\mathop{\tan }}\,}\frac{x}{2}\,dx}\] \[=\int{\frac{{{e}^{x}}{{\sec }^{2}}\frac{x}{2}}{2}}\,dx+{{e}^{x}}\tan \frac{x}{2}\] \[-\int{\frac{{{e}^{x}}{{\sec }^{2}}\frac{x}{2}}{2}}\,dx+K\] \[={{e}^{x}}\tan \frac{x}{2}+K\]
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