A) \[-{{e}^{x}}\tan \,\left( x/2 \right)\]
B) \[-{{e}^{x}}\cot \,\left( x/2 \right)\]
C) \[-\frac{1}{2}{{e}^{x}}\tan \,\left( \frac{x}{2} \right)\]
D) \[\frac{1}{2}{{e}^{x}}\cot \,\left( \frac{x}{2} \right)\]
Correct Answer: B
Solution :
\[I=\int{{{e}^{x}}\left( \frac{1-\sin x}{1-\cos x} \right)dx}\]\[=\int{{{e}^{x}}\left( \frac{1-\sin x}{2{{\sin }^{2}}(x/2)} \right)\,dx}\] Þ \[I=\int{{{e}^{x}}\left( \frac{1}{2}\text{cose}{{\text{c}}^{2}}\frac{x}{2}-\cot \frac{x}{2} \right)}\,dx\] \[\left( \because \,\,\int{{{e}^{x}}\left( f(x)+f'(x) \right)}={{e}^{x}}f(x)+c \right)\] \[\therefore \,\,I={{e}^{x}}\left( -\cot \frac{x}{2} \right)+c=-{{e}^{x}}\cot \frac{x}{2}+c\].
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