Question

If $A={\int }_0^{\pi }\frac{cosx}{(x+2{)}^2}dx,then{\int }_0^{\frac{\pi }{2}}\frac{sin2x}{(x+1)}$ dx is equal to

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

Answer 1

The correct option is B $\frac{1}{2}+\frac{1}{\pi +2}-A$

$A={\int }_0^{\pi }\frac{cosx}{(x+2{)}^2}dx={\left(cosx(-\frac{1}{x+2})\right)}_0^{\pi }-{\int }_0^{\pi }(-sinx)(-\frac{1}{x+2})dx$
$=\frac{1}{\pi +2}+\frac{1}{2}-{\int }_0^{\pi }\frac{sinx}{x+2}dx$
$A=\frac{1}{\pi +2}+\frac{1}{2}-2{\int }_0^{\frac{\pi }{2}}\frac{sin2x}{2x+2}dx$
${\int }_0^{\frac{\pi }{2}}\frac{sin2x}{1+x}dx=\frac{1}{\pi +2}+\frac{1}{2}-A$