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]^π/4_0 2∫_0^π/4tan θ d θ =π/2+2[log cos x]^π/4_0=π/4 2log √(2 ...

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Byju's Answer

Standard XIII

Mathematics

Property 1

∫01sin-1 2x/1...

Question

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

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

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Similar questions

1 \int 0 {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x

Q. Evaluate:

\int_{0}^{1} {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x

Q. Evalaute the integral

\int_{0}^{1} {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x

Evaluate the integrals using substitution.
$\int_{0}^{1} {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x$

Q. Evaluate the integral

\int_{0}^{1} {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x

using substitution.

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∫10sin−1(2x1+x2)dx=

The correct option is A π2−2 log√2Put x=tan θ, ∴ dx=sec2 θ dθ As x=1⇒θ=π4 and x=0⇒θ=0,then I=2∫π40θ sec2 θ d θ=2[θ tan θ]π40−2∫π40tan θ d θ =π2+2[log cos x]π40=π4−2log √2

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