Integral Calculus Question 106
Question: If $A=\int_0^\pi \frac{\cos x}{(x+2)^2} d x$, then $\int_0^{\frac{\pi}{2}} \frac{\sin (2 x)}{x+1} d x$ is equal to
Options:
A) $ \frac{1}{2}+\frac{1}{\pi +2}-A $
B) $ \frac{1}{\pi +2}-A $
C) $ 1+\frac{1}{\pi +2}-A $
D) $ A-\frac{1}{2}-\frac{1}{\pi +2} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ I=\int_0^{\pi /2}{\frac{\sin 2x}{x+1}dx.} $ Put $ x=y/2. $ Then, $ I=\int_0^{\pi }{\frac{\sin y}{y+2}dy} $ $ =( \frac{-\cos y}{y+2} )_0^{\pi }-\int\limits_0^{\pi }{\frac{\cos y}{{{(y+2)}^{2}}}dy} $ (Integrating by parts) $ =\frac{1}{\pi +2}+\frac{1}{2}-A $
BETA
