A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{12}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{8}\]
Correct Answer: C
Solution :
\[\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}\,x}{{{\sin }^{100}}\,x+\,{{\cos }^{100}}x}}\,\,dx\] \[=\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}}{{{\sin }^{100}}x+{{\sin }^{100}}\left( \frac{\pi }{2}-x \right)}}\,\,dx\] \[=\frac{\frac{\pi }{2}-0}{2}=\frac{\pi }{4}\] \[\left[ \because \,\,\int_{a}^{b}{\frac{f(x)\,dx}{f(x)+f(a+b-x)}=\frac{b-a}{2}} \right]\] Alternate \[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}x}{{{\sin }^{100}}x+{{\cos }^{100}}x}\,\,\,dx}\] ?. (i) \[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}\,\left( \frac{\pi }{2}-x \right)}{{{\sin }^{100}}\left( \frac{\pi }{2}-x \right)-{{\cos }^{100}}\left( \frac{\pi }{2}-x \right)}}\,\,\,dx\] \[\left[ \begin{align} & \because \,\,by\,\,define\,\,\text{integral property}\text{.} \\ & \int_{a}^{0}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \\ \end{align} \right]\] \[I=\int_{0}^{\pi /2}{\frac{{{\cos }^{100}}x}{{{\cos }^{100}}x+{{\sin }^{100}}x}}\,dx\] ?. (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{1\,\,dx\,=\frac{\pi }{2}-0\,\,\Rightarrow \,I=\frac{\pi }{4}}\]
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