-0- x sin x-1-cos2 xdx-

∫_0^π x sin x/1+cos^2 xdx= ∫_0^π (π x)sin x/1+cos^2 xdx= ∫_0^ππ sin x/1+cos^2 xdx ∫_0^πx sin x/1+cos^2xdx ⇒ 2I=π∫_0^πsin x/1+cos^2xdx=π [Tan^ 1(cos x)]^0_π = π ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

∫0π x sin x/1...

Question

$\int_{0}^{π} \frac{x s i n x}{1 + c o s^{2} x} d x =$

\frac{π^{2}}{2}

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\frac{π^{2}}{3}

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π^{2}

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\frac{π^{2}}{4}

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Solution

The correct option is D $\frac{π^{2}}{4}$
$\int_{0}^{π} \frac{x s i n x}{1 + c o s^{2} x} d x = \int_{0}^{π} \frac{(π - x) s i n x}{1 + c o s^{2} x} d x = \int_{0}^{π} \frac{π s i n x}{1 + c o s^{2} x} d x - \int_{0}^{π} \frac{x s i n x}{1 + c o s^{2} x} d x$
$\Rightarrow 2 I = π \int_{0}^{π} \frac{s i n x}{1 + c o s^{2} x} d x = π [T a n^{- 1} (c o s x)]_{π}^{0} = π [\frac{π}{4} + \frac{π}{4}] = \frac{π^{2}}{2} \Rightarrow I = \frac{π^{2}}{4}$

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∫π0 x sin x1+cos2 xdx=

The correct option is D π24∫π0 x sin x1+cos2 xdx=∫π0 (π−x)sin x1+cos2 xdx=∫π0π sin x1+cos2 xdx−∫π0x sin x1+cos2xdx ⇒2I=π∫π0sin x1+cos2xdx=π[Tan−1(cos x)]0π=π[π4+π4]=π22⇒I=π24

$\int_{0}^{π} \frac{x s i n x}{1 + c o s^{2} x} d x =$