Question

If $A={\int }_0^{\pi }\frac{cosx}{(x+2{)}^2}dx$, then ${\int }_0^{\frac{\pi }{2}}\frac{\sin 2x}{(x+1)}dx$ is equal to

• A. $\frac{1}{2}+\frac{1}{\pi +2}-A$
• B. $\frac{1}{\pi +2}-A$
• C. $1+\frac{1}{\pi +2}-A$
• D. $A-\frac{1}{2}-\frac{1}{\pi +2}$

Answer 1

Answer: (A)

$\frac{1}{2}+\frac{1}{\pi +2}-A$

${\int }_0^{\frac{\pi }{2}}\frac{\sin 2x}{(x+1)}dx$
Substitute $x=\frac{y}{2}$ Then,
$I={\int }_0^{\pi }\frac{\sin y}{y+2}dy$
$={\left(\frac{-\cos y}{y+2}\right)}_0^{\pi }-{\int }_0^{\pi }\frac{\cos y}{(y+2{)}^2}dy$
(Integrating by parts)
$=\frac{1}{\pi +2}+\frac{1}{2}-A$
Ans: A