A) \[{{e}^{\pi /4}}\log 2\]
B) \[-{{e}^{\pi /4}}\log 2\]
C) \[\frac{1}{2}{{e}^{\pi /4}}\log 2\]
D) \[-\frac{1}{2}{{e}^{\pi /4}}\log 2\]
Correct Answer: C
Solution :
Let \[I=\int_{\pi /4}^{\pi /2}{{{e}^{x}}(\log \sin x+\cot x)dx}\] \[I=\int_{\pi /4}^{\pi /2}{{{e}^{x}}\log \sin x\,dx+\int_{\pi /4}^{\pi /2}{{{e}^{x}}\cot x\,dx}}\] \[=\int_{\pi /4}^{\pi /2}{{{e}^{x}}\log \sin xdx+[{{e}^{x}}\log \sin x]_{\pi /4}^{\pi /2}}\] \[-\int_{\pi /4}^{\pi /2}{{{e}^{x}}\log \sin x\,dx}\] \[={{e}^{\pi /2}}\log \sin \frac{\pi }{2}-{{e}^{\pi /4}}\log \sin \frac{\pi }{4}=\frac{1}{2}{{e}^{\pi /4}}\log 2\].
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