-01x Sin-1 x-1-x2 d x-A- 1B- 1-2C- 0D- 2

The correct option is A 1Put θ=Sin^ 1x. Then d θ =1/√(1 x^2)dx. x=0, 1 ⇒θ =0, π/2 ∴∫_0^1 x Sin^ 1x/√(1 x^2)dx= ∫_0^π/2 sin θ d θ = [θ ( cos θ)]_0^π/2 ∫_0^π/ ...

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Standard XII

Mathematics

Property 1

∫01x Sin-1 x/...

Question

$\int_{0}^{1} \frac{x S i n^{- 1} x}{\sqrt{1 - x^{2}}} d x =$

0

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1

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\frac{1}{2}

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2

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Solution

The correct option is B $1$
$P u t θ = S i n^{- 1} x . T h e n d θ = \frac{1}{\sqrt{1 - x^{2}}} d x . x = 0, 1 \Rightarrow θ = 0, \frac{π}{2}$
$∴ \int_{0}^{1} \frac{x S i n^{- 1} x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{\frac{π}{2}} s i n θ d θ = [θ (- c o s θ)]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 (- c o s θ) d θ = [s i n θ]_{0}^{\frac{π}{2}} = 1$

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∫10 x Sin−1x√1−x2dx=

The correct option is B 1Put θ=Sin−1x. Then dθ=1√1−x2dx. x=0, 1⇒θ=0,π2 ∴∫10 x Sin−1x√1−x2dx=∫π20 sinθ dθ=[θ(−cosθ)]π20−∫π201(−cos θ)dθ=[sinθ]π20=1

$\int_{0}^{1} \frac{x S i n^{- 1} x}{\sqrt{1 - x^{2}}} d x =$