The correct option is A π 2/2 ∫_0^1 sin^ 1x dx ∫ sin^ 1x dx=x nbsp; sin^ 1x ∫x nbsp; dx /√(1 x^2) =x nbsp; sin^ 1x 1/2∫dx^2/√(1+x ...

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Special Integrals - 1

∫01Sin-1xdx=

Question

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

π - 2

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\frac{π - 2}{2}

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\frac{π + 2}{2}

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π + 2

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Solution

The correct option is A $\frac{π - 2}{2}$
$\int_{0}^{1} s i n^{- 1} x d x$
$\int s i n^{- 1} x d x = x s i n^{- 1} x - \int \frac{x d x}{\sqrt{1 - x^{2}}}$
$= x s i n^{- 1} x - \frac{1}{2} \int \frac{d x^{2}}{\sqrt{1 + x^{2}}} + c$
$= x s i n^{- 1} x + \frac{1}{2} \sqrt{1 - x^{2}} + c$
$= x s i n^{- 1} x + \sqrt{1 - x^{2}} + c$
$\int_{0}^{1} s i n^{- 1} x d x = (s i n^{- 1} 1 + 0 + c) - (0 + \sqrt{1} + c)$
$= \frac{π}{2} - 1$
$= \frac{π - 2}{2}$
$\int_{0}^{1} s i n^{- 1} x d x = \frac{π - 2}{2}$

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Similar questions

\int_{0}^{1} {sin}^{- 1} x d x = \frac{π}{2} - 1

Prove the following question.

$\int_{0}^{1} s i n^{- 1} x d x = \frac{π}{2} - 1.$

A = \int_{0}^{1} {sin}^{- 1} x d x

B = \int_{0}^{1} {cos}^{- 1} x d x

\frac{π}{2}

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

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

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