Mathematics

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

${\pi}^2$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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