Question

Let $I={\int }_0^{\pi }\frac{cosx}{(x+2{)}^2}dx$ and $J={\int }_0^{\frac{\pi }{2}}\frac{sin2x}{x+1}dx$, then the value of 'J' terms of 'I' equals

• A. J=2I
• B. J = 0
• C. $J=\frac{1}{\pi +2}+\frac{1}{2}-I$
  
• D. $J=\frac{1}{\pi +2}-\frac{1}{2}+I$

Answer 1

The correct option is C $J=\frac{1}{\pi +2}+\frac{1}{2}-I$


$J=2{\int }_0^{\frac{\pi }{2}}\frac{sin2x}{2x+2}dx,put2x=t$
Then $J={\int }_0^{\pi }\frac{sint}{t+2}dt=\left(\frac{1}{\pi +2}\right)+\frac{1}{2}-I$