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