Question
$$\int_0^{\frac{\pi }{2}} {\frac{{f\left( x \right)}}{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}dx,} $$ where $$f\left( x \right) \ne - f\left( {\frac{\pi }{2} - x} \right)$$ for $$0 \leqslant x \leqslant \frac{\pi }{2},$$ has the value :
A.
$$f\left( 0 \right)$$
B.
$$f\left( {\frac{\pi }{2}} \right)$$
C.
$$\frac{\pi }{2}$$
D.
none of these
Answer :
none of these
Solution :
$$\eqalign{
& I = \int_0^{\frac{\pi }{2}} {\frac{{f\left( x \right)dx}}{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}} \cr
& \,\,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{f\left( {\frac{\pi }{2} - x} \right)}}{{f\left( {\frac{\pi }{2} - x} \right) + f\left( x \right)}}dx} \cr
& \left( {{\text{using }}\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} } \right) \cr
& \therefore I + I = \int_0^{\frac{\pi }{2}} {\frac{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}} dx\, = \int_0^{\frac{\pi }{2}} {dx} \, = \left[ x \right]_0^{\frac{\pi }{2}}\, = \frac{\pi }{2} \cr
& \therefore I = \frac{\pi }{4} \cr} $$