A) A sq unit
B) \[\frac{\pi }{2}-A\] sq unit
C) \[-A\] sq unit
D) None of these
Correct Answer: A
Solution :
Given, \[\int_{0}^{\pi /2}{\sin xdx=A}\] ...(i) Now, \[\int_{0}^{\pi /2}{\cos xdx}\] \[=\int_{0}^{\pi /2}{\cos \left( \frac{\pi }{2}-x \right)dx}\] \[=\int_{0}^{\pi /2}{(\sin x)dx}\] \[=\int_{0}^{\pi /2}{\sin xdx}\] \[=A\] [from Eq.(i)]
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