Integral Calculus Question 205
Question: What is $ \int\limits_0^{\pi /2}{\sin 2x\ell n,(\cot x)dx} $ equal to?
Options:
A) 0
B) $ \pi \ell n2 $
C) $ -\pi \ell n2 $
D) $ \frac{\pi \ell n2}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ I=\int\limits_0^{\frac{\pi }{2}}{\sin 2x\ell n(\cot x)dx} $ $ =\int\limits_0^{\frac{\pi }{2}}{\sin 2x\ell n(\cos x)dx-\int\limits_0^{\frac{\pi }{2}}{\sin 2x\ell n(\sin ,x)dx}} $ $ \int\limits_0^{\frac{\pi }{2}}{\sin [ 2( \frac{\pi }{2}+x ) ]\ell n,\cos ( \frac{\pi }{2}+x )dx} $ $ -\int\limits_0^{\frac{\pi }{2}}{\sin 2x,\ell n(\sin x)dx} $ $ =\int\limits_0^{\frac{\pi }{2}}{\sin (\pi +2x)\ell n(\sin ,x)dx} $ $ -\int\limits_0^{\frac{\pi }{2}}{\sin 2x\ell n(\sin ,x)dx} $ $ =\int\limits_0^{\frac{\pi }{2}}{\sin 2x\ell n}(\sin x)dx-\int\limits_0^{\frac{\pi }{2}}{\sin 2x\ell n(\sin x)dx} $ $ =0 $
BETA
