A) \[{{I}_{1}}+{{I}_{2}}=\frac{\pi }{2}\]
B) \[{{I}_{2}}-{{I}_{1}}=\frac{\pi }{2}\]
C) \[{{I}_{1}}+{{I}_{2}}=0\]
D) \[{{I}_{1}}={{I}_{2}}\]
Correct Answer: B
Solution :
Since, \[{{I}_{1}}-\int_{0}^{\pi /2}{x\sin }x\,dx\] and \[{{I}_{2}}-\int_{0}^{\pi /2}{x\cos }x\,dx\] \[\therefore \] \[{{I}_{1}}=\int_{0}^{\pi /2}{x\,\sin x\,dx}\] \[\Rightarrow \] \[{{I}_{1}}=-[x\cos x]_{0}^{\pi /2}-\int_{0}^{\pi /2}{(-\cos x)}dx\] \[=[0-0]+[\sin x]_{0}^{\pi /2}\] \[=1\] ?(i) and \[{{I}_{2}}=\int_{0}^{\pi /2}{x\cos x\,dx}\] \[=[x\sin x]_{0}^{\pi /2}-\int_{0}^{\pi /2}{\cos xdx}\] \[=\left( \frac{\pi }{2}-0 \right)+[x\sin x]_{0}^{\pi /2}\] \[=\frac{\pi }{2}+1\] ?(i) From Eqs. (i) and (ii) \[{{I}_{2}}-{{I}_{1}}=\frac{\pi }{2}\]
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