A) \[{{\pi }^{2}}\]
B) \[2\pi \]
C) \[2{{\pi }^{2}}\]
D) \[4\pi \]
Correct Answer: A
Solution :
[a] Let \[I=\int_{0}^{2\pi }{\frac{x.{{\sin }^{8}}x}{{{\sin }^{8}}x+{{\cos }^{8}}x}}\,\,dx\] ?.(i) \[I=\int_{0}^{2\pi }{\frac{(2\pi -x)si{{n}^{8}}x}{{{\sin }^{8}}x+{{\cos }^{8}}x}}\,\,dx\] ?.(ii) \[(i)+(ii)\] \[2I=2\pi \int_{0}^{2\pi }{\frac{{{\sin }^{8}}x}{{{\sin }^{8}}x+{{\cos }^{8}}x}}\,\,dx\] \[\Rightarrow \,\,\,\,\,\,I=4\pi \int_{0}^{\pi /2}{\frac{{{\sin }^{8}}x}{{{\sin }^{8}}x+{{\cos }^{8}}x}}\,dx\]?..(iii) Again \[I=4\pi \int_{0}^{\pi /2}{\frac{{{\cos }^{8}}x}{{{\sin }^{8}}x+{{\cos }^{8}}x}}\,\,dx\] ...(iv) \[(iii)+(iv)\] \[2I=4\pi \int_{0}^{\pi /2}{dx}=2{{\pi }^{2}}\Rightarrow I={{\pi }^{2}}\]
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