Definite Integration Question 363
Question: $ \int _{\pi \text{/4}}^{\pi \text{/2}}{e^{x}(\log \sin x+\cot x)dx=} $
[AI CBSE 1991]
Options:
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 $
Show Answer
Answer:
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 xdx+\int _{\pi /4}^{\pi /2}{e^{x}\cot xdx}} $
$ =\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 xdx} $
$ ={e^{\pi /2}}\log \sin \frac{\pi }{2}-{e^{\pi /4}}\log \sin \frac{\pi }{4}=\frac{1}{2}{e^{\pi /4}}\log 2 $ .
BETA
