If -0xx-1-sin x2 d x- then -0-2 x2cos 2x-2-1-sin x2 d x isA- -2 -2B- -4 -2- -D- -

The correct option is A λ + 2 π π^2 I = ∫^π_0 2x^2 cos^2 x/2/(1 + sin x)^2 dx = ∫^π_0 x^2 (cos x + 1)/(1 + sin x)^2dx = ∫^π_0 x^2/(1 + sin x)^2dx + = ∫^π_ ...

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Standard XII

Mathematics

Property 7

If ∫0xx/1+sin...

Question

If $\int_{0}^{x} (\frac{x}{1 + s i n x})^{2} d x = λ, t h e n \int_{0}^{π} \frac{2 x^{2} c o s^{2} \frac{x}{2}}{(1 + s i n x)^{2}} d x$ is

$λ$

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$λ + 2 π - π^{2}$

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$λ - π$

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$λ + 4 π - π^{2}$

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Solution

The correct option is B
$λ + 2 π - π^{2}$

$I = \int_{0}^{π} \frac{2 x^{2} c o s^{2} \frac{x}{2}}{(1 + s i n x)^{2}} d x = \int_{0}^{π} \frac{x^{2} (c o s x + 1)}{(1 + s i n x)^{2}} d x = \int_{0}^{π} \frac{x^{2}}{(1 + s i n x)^{2}} d x + = \int_{0}^{π} \frac{x^{2} c o s x}{(1 + s i n x)^{2}} d x = λ + = \int_{0}^{π} \frac{x^{2} c o s x}{(1 + s i n x)^{2}} d x L e t I_{1} == \int_{0}^{π} x_{1}^{2} \frac{c o s x}{(1 + s i n x)^{2} I I} d x = x^{2} \int \frac{c o s x}{(1 + s i n x)^{2}} d x + \int \frac{2 x}{(1 + s i n x)} d x]_{0}^{π} = - π + 2 π ∴ I = λ - π^{2} + 2 π$

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If ∫x0(x1+sin x)2dx=λ,then∫π02x2cos2x2(1+sin x)2dx is

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If $\int_{0}^{x} (\frac{x}{1 + s i n x})^{2} d x = λ, t h e n \int_{0}^{π} \frac{2 x^{2} c o s^{2} \frac{x}{2}}{(1 + s i n x)^{2}} d x$ is