A) \[1\]
B) \[0\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{4}\]
E) \[\pi \]
Correct Answer: D
Solution :
Let\[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{n}}\theta }{{{\sin }^{n}}\theta +{{\cos }^{n}}\theta }}d\theta \] ?.(i) \[=\int_{0}^{\pi /2}{\frac{{{\sin }^{n}}(\pi /2-\theta )}{{{\sin }^{n}}(\pi /2-\theta )+{{\cos }^{n}}(\pi /2-\theta )}}d\theta \] \[=\int_{0}^{\pi /2}{\frac{{{\cos }^{n}}\theta }{{{\cos }^{n}}\theta +{{\sin }^{n}}\theta }}d\theta \] ?(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{\frac{{{\sin }^{n}}\theta +{{\cos }^{n}}\theta }{{{\cos }^{n}}\theta +{{\sin }^{n}}\theta }}d\theta \] \[=\int_{0}^{\pi /2}{d\theta =\frac{\pi }{2}}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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