The correct option is A #160; x( sin^ 1x)^2 + 2( sin^ 1x)√(1 x^2) #160; 2x + C. [ ...

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Differentiation of Inverse Trigonometric Functions

∫ sin - 1x2dx

Question

$\int {({sin}^{- 1} x)}^{2} d x$

x {({sin}^{- 1} x)}^{2} + 2 ({sin}^{- 1} x) \sqrt{1 - x^{2}} - 2 x + C .

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- 2 t a n^{- 1} \sqrt{1 - t a n x + c o t x + c}

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2 ({sin}^{- 1} x) \sqrt{1 - x^{2}} - 2 x + C

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- 2 t a n^{- 1} \sqrt{t a n x + c o t x - 1 + c}

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Solution

The correct option is A $x {({sin}^{- 1} x)}^{2} + 2 ({sin}^{- 1} x) \sqrt{1 - x^{2}} - 2 x + C .$
$\begin{matrix} W e h a v e \int {({sin}^{- 1} x)}^{2} d x p u t t i n g x = sin t a n d d x = cos t d t n o w, w e g e t \int {({sin}^{- 1} x)}^{2} d x = \int t^{2} cos t d t = t^{2} . sin (t) - \int 2 t (sin t) d t = t^{2} sin t - 2 [t (- cos t) - \int 1. (- cos t) d t] = t^{2} sin t + 2 t cos t - 2 sin t + C = x {({sin}^{- 1} x)}^{2} + 2 ({sin}^{- 1} x) \sqrt{1 - x^{2}} - 2 x + C . H e n c e, t h e o p t i o n A i s c o r r e c t . \end{matrix}$

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Similar questions

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

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{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

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π + {t a n}^{- 1} \frac{x + y}{1 - x y}

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3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

\int {sin}^{- 1} x {cos}^{- 1} x d x = f^{- 1} (x) [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}] \frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C

$\int \frac{x^{2} + 1}{x^{4} + 1} d x$ will be equal to which of the following

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