Question
The value of $$\int_0^{\frac{\pi }{2}} {\frac{{dx}}{{1 + {{\tan }^3}x}}} $$ is :
A.
$$\frac{\pi }{2}$$
B.
$$\frac{\pi }{4}$$
C.
$$\pi $$
D.
none of these
Answer :
$$\frac{\pi }{4}$$
Solution :
$$\eqalign{
& I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x}}{{{{\cos }^3}x + {{\sin }^3}x}}dx} \cr
& \,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}\left( {\frac{\pi }{2} - x} \right)}}{{{{\cos }^3}\left( {\frac{\pi }{2} - x} \right) + {{\sin }^3}\left( {\frac{\pi }{2} - \pi } \right)}}dx} \cr
& \,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx} \cr
& \therefore I + I = \int_0^{\frac{\pi }{2}} {1\,dx} = \frac{\pi }{2} \cr
& \therefore I = \frac{\pi }{4} \cr} $$