A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[2\pi \]
Correct Answer: A
Solution :
Let \[I=\int_{0}^{\pi /2}{\frac{\sqrt[3]{{{\sin }^{2}}x}}{\sqrt[3]{{{\sin }^{2}}x}+\sqrt[3]{{{\cos }^{2}}x}}}dx\] ...(i) \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sqrt[3]{{{\sin }^{2}}(\pi /2-x)}}{\sqrt[3]{{{\sin }^{2}}(\pi /2-x)}+\sqrt[3]{{{\cos }^{2}}(\pi /2-x)}}}dx\] \[=\int_{0}^{\pi /2}{\frac{\sqrt[3]{{{\cos }^{2}}x}}{\sqrt[3]{{{\cos }^{2}}x}+\sqrt[3]{{{\sin }^{2}}x}}}dx\] ?.(ii) On adding Eqs. (i) and (ii), \[2I=\int_{00}^{\pi /2}{1dx}\] \[\Rightarrow \] \[2I=[x]_{0}^{\pi /2}=\frac{\pi }{2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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