Mathematics

$$\displaystyle \int _{ 0 }^{ 2\pi  }{ \dfrac { x\sin ^{ 2n }{ x }  }{ \sin ^{ 2n }{ x } +\cos ^{ 2n }{ x }  }  } dx$$.


ANSWER

$${\pi}^2$$


SOLUTION
$$I = \int_0^{2\pi } {\dfrac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx$$
$$\Rightarrow I = \int_0^{2\pi } {\dfrac{{\left( {2\pi  - x} \right){{\sin }^{2n}}\left( {2\pi  - x} \right)}}{{{{\sin }^{2n}}\left( {2\pi  - x} \right) + {{\cos }^{2n}}\left( {2\pi  - x} \right)}}} dx$$     ($$\because \int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} } $$)
$$\Rightarrow I = \int_0^{2\pi } {\dfrac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx - \int_0^{2\pi } {\dfrac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$
$$ \Rightarrow I= \int_0^{2\pi } {\dfrac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx - I$$
$$\Rightarrow 2I=$$$$\int_0^{2\pi } {\dfrac{{2\pi {{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx$$
$$ \Rightarrow 2I = 2\pi \int_0^{2\pi } {\dfrac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$
$$ \Rightarrow I = 2\pi \int_0^\pi  {\dfrac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx$$   ($$\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^{2a} {f\left( x \right)dx} } $$)
$$ \Rightarrow I = 4\pi \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $$  ---(i)      ($$\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^{2a} {f\left( x \right)dx} } $$)
$$ \Rightarrow I = 4\pi \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^{2n}}x}}{{{{\cos }^{2n}}x + {{\sin }^{2n}}x}}} dx{\text{   }}$$    ---(ii)      ($$\because \int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} } $$)
Adding (i) and (ii) we get 
$$ \Rightarrow 2I = 4\pi \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx{\text{  }}$$

$$ \Rightarrow I = 2\pi \int_0^{\cfrac{\pi }{2}} {dx} $$

$$ \Rightarrow I = 2\pi \left[ x \right]_0^{\cfrac{\pi }{2}}$$

$$ \Rightarrow I = 2\pi  \times \dfrac{\pi }{2}$$

$$ \Rightarrow I = {\pi ^2}$$
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Single Correct Medium Published on 17th 09, 2020
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