The value of the integral ∫0π xsin⁡x1+cos2⁡xdx, is

A
$\frac{π^{2}}{2}$
B
$\frac{π^{2}}{4}$
C
$\frac{π^{2}}{8}$
D
$\frac{π^{2}}{16}$

Solution:

Let $I = \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x$

Then,

$\begin{array}{l} \Rightarrow I = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 - \cos^{2} (π - x)} d x \\ \Rightarrow I = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} d x \end{array}$

Adding (i) and (ii), we get

$\begin{array}{l} 2 π = π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} d x \\ \Rightarrow 2 I = - π \int_{1}^{- 1} \frac{1}{1 + t^{2}} d t, where t = \cos x \\ \Rightarrow 2 I = - π {[\tan^{- 1} t]}_{1}^{- 1} = - π (- \frac{π}{4} - \frac{π}{4}) = \frac{π^{2}}{2} \\ \Rightarrow I = \frac{π^{2}}{4} \end{array}$

The value of the integral $\int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x$ , is

Solution:

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