If \[A=\int_{0}^{\pi }{\frac{\operatorname{cosx}}{{{(x+2)}^{2}}}dx,}\] then \[A=\int_{0}^{\pi /2}{\frac{\sin 2x}{x+1}dx,}\] is equal to

\[\frac{1}{2}+\frac{1}{\pi +2}-A\]

\[A-\frac{1}{2}-\frac{1}{\pi +2}\]

If \[A=\int_{0}^{\pi }{\frac{\operatorname{cosx}}{{{(x+2)}^{2}}}dx,}\]

1.	If \[A=\int_{0}^{\pi }{\frac{\operatorname{cosx}}{{{(x+2)}^{2}}}dx,}\] then \[A=\int_{0}^{\pi /2}{\frac{\sin 2x}{x+1}dx,}\] 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» B. \[\frac{1}{\pi +2}-A\]

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If \[A=\int_{0}^{\pi }{\frac{\operatorname{cosx}}{{{(x+2)}^{2}}}dx,}\] then \[A=\int_{0}^{\pi /2}{\frac{\sin 2x}{x+1}dx,}\] is equal to

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