For n > 0, _0^ 2 x ^ 2n x ^ 2n x + ^ 2n x ,dx =

MCQ

For $n > 0,$ $\int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}\,dx = } $

A
${\pi ^{ - 1}}$
B
$\pi $
C
${\pi ^{ - 2}}$
✓
${\pi ^2}$

✓

Answer

Correct option: D.

${\pi ^2}$

d
(d) Let $I = \int_0^{2\pi } {\frac{{x{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
$ = \int_0^{2\pi } {\frac{{(2\pi - x){{[\sin (2\pi - x)]}^{2n}}}}{{{{[\sin (2\pi - x)]}^{2n}} + {{[\cos (2\pi - x)]}^{2n}}}}dx} $,
$[\because \,\int_0^a {f(x)dx = \int_0^a {f(a - x)\,dx} ]} $
$ = \int_0^{2\pi } {\frac{{(2\pi - x){{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
==> $2I = 2\pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
==> $I = \pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
Using the property $\int_0^{2a} {f(x)\,dx = \int_0^a {[f(x) + f(2a - x)]} \,dx} $
$I = 2\pi \int_0^\pi {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
Using the above property once again, we get
$I = 4\pi \int_0^{\pi /2} {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $.....$(i)$
==> $I = 4\pi \int_0^{\pi /2} {\frac{{{{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $ .....$(ii)$

Now, adding $(i)$ and $(ii),$ we get
$2I = 4\pi \int_0^{\pi /2} {\frac{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} $
$ = 4\pi \int_0^{\pi /2} {dx = (4\pi )\left( {\frac{\pi }{2}} \right) = 2{\pi ^2} \Rightarrow I = {\pi ^2}} $.

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