Let I=I↓∫sin^ 1x.II↓1dx =sin^ 1x· x ∫[1√(1 x^2)· x ]dx Let nbsp; I_1=∫x√(1 x^2)dx Put nbsp; 1 x^2=t^2 2xdx=2tdt⇒ xdx= tdt I_1=∫ ttdt I_1= t I_1= √(1 x^2) ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 1

∫sin-1 xdx=

Question

$\int {sin}^{- 1} x d x =$

x {sin}^{- 1} x + \sqrt{x^{2} - 1} + c

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x {sin}^{- 1} x - \sqrt{x^{2} - 1} + c

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x {sin}^{- 1} x + \sqrt{1 - x^{2}} + c

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x {sin}^{- 1} x - \sqrt{1 - x^{2}} + c

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Solution

The correct option is C $x {sin}^{- 1} x + \sqrt{1 - x^{2}} + c$

Let $I = \int {sin}^{- 1} x . ↓ I 1 d x ↓ I I$
$= {sin}^{- 1} x \cdot x - \int [\frac{1}{\sqrt{1 - x^{2}}} \cdot x] d x$
Let $I_{1} = \int \frac{x}{\sqrt{1 - x^{2}}} d x$
Put $1 - x^{2} = t^{2}$
$- 2 x d x = 2 t d t \Rightarrow x d x = - t d t$
$I_{1} = \int \frac{- t}{t} d t$
$I_{1} = - t$
$I_{1} = - \sqrt{1 - x^{2}}$
$I = x {sin}^{- 1} x + \sqrt{1 - x^{2}} + c$

Suggest Corrections

∫sin−1xdx=

The correct option is C xsin−1 x+√1−x2+c Let I=∫sin−1x.↓I1dx↓II =sin−1x⋅x−∫[1√1−x2⋅x]dx Let I1=∫x√1−x2dx Put 1−x2=t2 −2xdx=2tdt⇒xdx=−tdt I1=∫−ttdt I1=−t I1=−√1−x2 I=xsin−1x+√1−x2+c

$\int {sin}^{- 1} x d x =$