Question
$$\int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx,\,n > 0} $$ is equal to :
A.
$$\pi $$
B.
$$2\pi $$
C.
$${\pi ^2}$$
D.
$$\frac{1}{2}{\pi ^2}$$
Answer :
$${\pi ^2}$$
Solution :
$$\eqalign{
& I = \int_0^{2\pi } {\frac{{\left( {2\pi - x} \right){{\sin }^{2n}}\left( {2\pi - x} \right)}}{{{{\sin }^{2n}}\left( {2\pi - x} \right) + {{\cos }^{2n}}\left( {2\pi - x} \right)}}dx} \cr
& \,\,\,\,\, = 2\pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx - I} \cr
& \therefore 2I = 2\pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\pi .2\int_0^\pi {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \cr} $$
( because the integrand is a periodic function of period $$\pi .$$ )
$$\eqalign{
& I = 2\pi \int_{\frac{\pi }{2}}^{ - \frac{\pi }{2}} {\frac{{{{\sin }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}{{{{\sin }^{2n}}\left( {\frac{\pi }{2} - z} \right) + {{\cos }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}} d\left( {\frac{\pi }{2} - z} \right) \cr
& \,\,\,\,\, = 2\pi \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}z}}{{{{\cos }^{2n}}z + {{\sin }^{2n}}z}}} dz \cr
& \,\,\,\,\, = 4\pi \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}z}}{{{{\cos }^{2n}}z + {{\sin }^{2n}}z}}dz} \cr
& \therefore I = 4\pi I',\,{\text{where}}\,I' = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}z}}{{{{\cos }^{2n}}z + {{\sin }^{2n}}z}}dz} . \cr
& \therefore I' = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}{{{{\cos }^{2n}}\left( {\frac{\pi }{2} - z} \right) + {{\sin }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}} dz \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{2n}}z}}{{{{\sin }^{2n}}z + {{\cos }^{2n}}z}}} dz \cr
& \therefore I' + I' = \int_0^{\frac{\pi }{2}} {dz} = \frac{\pi }{2}\,\,\,\,\,\therefore I' = \frac{\pi }{4} \cr
& \therefore I = 4\pi I'\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow I = {\pi ^2} \cr} $$