A) \[-{{e}^{x}}.cot\frac{x}{2}+c\]
B) \[{{e}^{x}}.\tan +c\]
C) \[{{e}^{x}}.\cot \frac{x}{2}+c\]
D) \[-{{e}^{x}}.\tan \frac{x}{2}+c\]
Correct Answer: A
Solution :
[a] \[I=\int{{{e}^{x}}\left( \frac{1-\sin x}{1-\cos x} \right)}.dx\] \[=\int{{{e}^{x}}\left( \frac{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}-2.\sin \frac{x}{2}.\cos \frac{x}{2}}{2{{\sin }^{2}}\frac{x}{2}} \right)}.dx\frac{1}{2}{{\int{{{e}^{x}}\left( \frac{\sin \frac{x}{2}-\cos \frac{x}{2}}{\sin \frac{x}{2}} \right)}}^{2}}dx\] \[=\frac{1}{2}\int{{{e}^{x}}{{\left( 1-\cot \frac{x}{2} \right)}^{2}}dx=\frac{1}{2}\int{{{e}^{x}}\left( 1+{{\cot }^{2}}\frac{x}{2}-2\cot \frac{x}{2} \right)}.dx}\]Here \[f\left( x \right)=-2cot\frac{x}{2}\] \[\therefore I=\frac{1}{2}.{{e}^{x}}.\left( -2\cot \frac{x}{2} \right)+c=-{{e}^{x}}\cot \frac{x}{2}+c\] Hence, option [a] is correct.
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