∫02π xsin2n⁡xsin2n⁡x+cos2n⁡xdx,n>0, is equal to

A
$π$
B
$2 π$
C
$π^{2}$
D
$\frac{1}{2} π^{2}$

Solution:

Let $I = \int_{0}^{2 π} \frac{x \sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x$ …(i)

Using property IV, we have

$\begin{array}{l} I_{n} = \int_{0}^{2 π} \frac{(2 π - x) \sin^{2 n} (2 π - x)}{\sin^{2 n} (2 π - x) + \cos^{2 n} (2 π - x)} d x \\ \Rightarrow I_{n} = \int_{0}^{2 π} \frac{(2 π - x) \sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x \end{array}$

Adding (i) and (ii), we get

$\begin{array}{l} 2 I_{n} = 2 π \int_{0}^{2 π} \frac{\sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x \\ \Rightarrow I_{n} = π \int_{0}^{2 π} \frac{\sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} ∣ x} d x \\ \Rightarrow I_{n} = 2 π \int_{0}^{π} \frac{\sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x [Using property VII] \\ \Rightarrow I_{n} = 4 π \int_{0}^{π / 2} \frac{\sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x [Using property VII] \\ \Rightarrow I_{n} = 4 π \times \frac{π}{4} = π^{2} \end{array}$

$\int_{0}^{2 π} \frac{x \sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x, n > 0,$ is equal to

Solution:

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