A) \[{{I}_{1}}=2{{I}_{2}}\]
B) \[2{{I}_{1}}={{I}_{2}}\]
C) \[{{I}_{1}}={{I}_{2}}\]
D) \[{{I}_{1}}+{{I}_{2}}=0\]
Correct Answer: C
Solution :
[c] Consider \[{{I}_{1}}=\int\limits_{0}^{\pi }{xf\left[ {{\sin }^{3}}x+{{\cos }^{2}}x \right]}dx\] \[=\int\limits_{0}^{\pi }{(\pi -x)f\left[ {{\sin }^{3}}(\pi -x)+{{\cos }^{2}}(\pi -x) \right]dx}\] \[=\int\limits_{0}^{\pi }{(\pi -x)f\left[ {{\sin }^{3}}x+{{\cos }^{2}}x \right]dx}\] \[=\int\limits_{0}^{\pi }{\pi f({{\sin }^{3}}x+{{\cos }^{2}}x)dx-\int\limits_{0}^{\pi }{xf({{\sin }^{3}}x+{{\cos }^{2}}x)dx}}\] \[=\int\limits_{0}^{\pi }{\pi f({{\sin }^{3}}x+{{\cos }^{2}}x)dx-{{I}_{1}}}\] \[\Rightarrow 2{{I}_{1}}=\int\limits_{0}^{\pi }{\pi f({{\sin }^{3}}x+{{\cos }^{2}}x)dx}\] \[2{{I}_{1}}=2\pi \int\limits_{0}^{\pi /2}{f({{\sin }^{3}}x+{{\cos }^{2}}x)dx}\] \[\Rightarrow {{I}_{1}}=\pi \int\limits_{0}^{\pi /2}{f({{\sin }^{3}}x+{{\cos }^{2}}x)dx}\] \[{{I}_{1}}={{I}_{2}}\] (By definition of \[{{I}_{2}}\])
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