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 \[I=\int_{{}}^{{}}{\left( \frac{2-\sin 2x}{1-\cos \,2x} \right)\,{{e}^{x}}dx}\] \[I=\int_{{}}^{{}}{\left( \frac{2-2\sin x\cos x}{2{{\sin }^{2}}x} \right){{e}^{x}}dx}\] \[=\int_{{}}^{{}}{\underset{II}{\mathop{{{\operatorname{cosec}}^{2}}}}\,x\underset{I}{\mathop{{{e}^{x}}}}\,dx-\int_{{}}^{{}}{\cot \,x\,{{e}^{x}}dx}}\] \[=-\cot \,x\,{{e}^{x}}-\int_{{}}^{{}}{(-\cos \,x){{e}^{x}}\,dx}\] \[-\int_{{}}^{{}}{\cot \,x{{e}^{x}}\,dx\,+c}\] \[=-\cot \,x\,{{e}^{x}}\,+c\]You need to login to perform this action.
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