-8747-x -160-sin-1x1-x2dx-1- x-1-x2 -160-sin-1x-c nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -2- x-sin-1x-c-3- x-1-x2 -160-sin-1x-c nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -4- x22sin-1x-1-x2 -160-c

Dear student I= #8747;xsin 1x1 x2dxIntegrate #160;by #160;parts: #160;f=sin 1x,g'=x1 x2,f'=11 x2,g= 1 x2So, #160;I= 1 x2sin 1x #8747; dx= 1 x2sin 1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiation Using Substitution

#8747;x #16...

Question

$\int \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x =$
(1) $x + \sqrt{1 - x^{2}} \sin^{- 1} x + c$ (2) $x - \sin^{- 1} x + c$
(3) $x - \sqrt{1 - x^{2}} \sin^{- 1} x + c$ (4) $\frac{x^{2}}{2} \sin^{- 1} x - \sqrt{1 - x^{2}} + c$

Open in App

Solution

Dear student
$I = \int \frac{{xsin}^{- 1} x}{\sqrt{1 - x^{2}} dx} Integrate by parts : f = \sin^{- 1} x, g' = \frac{x}{\sqrt{1 - x^{2}}}, f' = \frac{1}{\sqrt{1 - x^{2}}}, g = - \sqrt{1 - x^{2}} So, I = - \sqrt{1 - x^{2}} \sin^{- 1} x - \int - dx = - \sqrt{1 - x^{2}} \sin^{- 1} x + x + C$
Regards

Suggest Corrections

Similar questions

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

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

x y < 1

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

x y > 1

.
(a)

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}

(b)

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

(c)

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

Q. Solve for

x

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

Q. The solution of the equation

{sin}^{- 1} \sqrt{1 - x^{2}} + {cos}^{- 1} x = {cot}^{- 1} \frac{\sqrt{1 - x^{2}}}{x} - {sin}^{- 1} x

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

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

x y < 1

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

x y > 1

.
If

0 < x < 1

, then

\sqrt{1 + x^{2}} {[{x c o s ({c o t}^{- 1} x) + s i n ({c o t}^{- 1} x)}^{2} - 1]}^{1 / 2}

is equal to
(a)

\frac{x}{\sqrt{1 + x^{2}}}

(b)

x

(c)

x \sqrt{1 + x^{2}}

(d)

\sqrt{1 + x^{2}}

Q. The solution set of the equation

{sin}^{- 1} \sqrt{1 - x^{2}} + {cos}^{- 1} x = {cot}^{- 1} \frac{\sqrt{1 - x^{2}}}{x} - {sin}^{- 1} x

is?

Join BYJU'S Learning Program

∫x sin-1x1-x2dx= (1) x+1-x2 sin-1x+c (2) x-sin-1x+c (3) x-1-x2 sin-1x+c (4) x22sin-1x-1-x2 +c

Dear student I=∫xsin-1x1-x2dxIntegrate by parts: f=sin-1x,g'=x1-x2,f'=11-x2,g=-1-x2So, I=-1-x2sin-1x-∫-dx=-1-x2sin-1x+x+C Regards

$\int \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x =$
(1) $x + \sqrt{1 - x^{2}} \sin^{- 1} x + c$ (2) $x - \sin^{- 1} x + c$
(3) $x - \sqrt{1 - x^{2}} \sin^{- 1} x + c$ (4) $\frac{x^{2}}{2} \sin^{- 1} x - \sqrt{1 - x^{2}} + c$

Dear student
$I = \int \frac{{xsin}^{- 1} x}{\sqrt{1 - x^{2}} dx} Integrate by parts : f = \sin^{- 1} x, g' = \frac{x}{\sqrt{1 - x^{2}}}, f' = \frac{1}{\sqrt{1 - x^{2}}}, g = - \sqrt{1 - x^{2}} So, I = - \sqrt{1 - x^{2}} \sin^{- 1} x - \int - dx = - \sqrt{1 - x^{2}} \sin^{- 1} x + x + C$
Regards