-01 sin-1 2x-1-x2 dx - -Karnataka CET 1999-

Put x = tan θ, ∴ dx = sec^2 θ d θ As x=1 ⇒θ =π/4 and x = 0 ⇒θ = 0, then I=2 ∫_0^π/4θ sec^2 θ d θ = 2[θ tan θ]_0^π/4 2 ∫_0^π/4 tan θ d θ =π/2+2 [log cos x]^π/4 ...

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Standard XIII

Mathematics

Property 1

∫01 sin-1 2x/...

Question

$\int_{0}^{1} s i n^{- 1} (\frac{2 x}{1 + x^{2}}) d x =$ [Karnataka CET 1999]

\frac{π}{2} - 2 l o g \sqrt{2}

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\frac{π}{2} + 2 l o g \sqrt{2}

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\frac{π}{4} - l o g \sqrt{2}

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\frac{π}{4} + l o g \sqrt{2}

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Solution

The correct option is A $\frac{π}{2} - 2 l o g \sqrt{2}$
$P u t x = t a n θ, ∴ d x = s e c^{2} θ d θ$
$A s x = 1 \Rightarrow θ = \frac{π}{4} a n d x = 0 \Rightarrow θ = 0, t h e n$
$I = 2 \int_{0}^{\frac{π}{4}} θ s e c^{2} θ d θ = 2 [θ t a n θ]_{0}^{\frac{π}{4}} - 2 \int_{0}^{\frac{π}{4}} t a n θ d θ$
$= \frac{π}{2} + 2 [l o g c o s x]_{0}^{\frac{π}{4}} = \frac{π}{2} - 2 l o g \sqrt{2}$

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