-01-2x sin-1x-1-x2dx-

Put t=sin^ 1x ⇒ dt=1/√(1 x^2)dx, then ∫_0^1/2x sin^ 1x/√(1 x^2)dx= ∫_0^π/6 t sin t dt=[ t cos t+sin t]_0^π/6 = [ π/6.√(3)/2+1/2 ]= [ 1/2 √(3)π/12 ] ...

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

Byju's Answer

Standard XI

Mathematics

Property 1

∫01/2x sin-1x...

Question

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

\frac{1}{2} + \frac{\sqrt{3} π}{12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} - \frac{\sqrt{3} π}{12}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{2} - \frac{\sqrt{3} π}{12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{2} - \frac{\sqrt{3} π}{12}$
$P u t t = s i n^{- 1} x \Rightarrow d t = \frac{1}{\sqrt{1 - x^{2}}} d x, t h e n$
$\int_{0}^{\frac{1}{2}} \frac{x s i n^{- 1} x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{\frac{π}{6}} t s i n t d t = [- t c o s t + s i n t]_{0}^{\frac{π}{6}}$
$= [- \frac{π}{6} . \frac{\sqrt{3}}{2} + \frac{1}{2}] = [\frac{1}{2} - \frac{\sqrt{3} π}{12}]$

Suggest Corrections

Similar questions

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

\int_{0}^{1} \frac{x S i n^{- 1} x}{\sqrt{1 - x^{2}}} d x =

Q. Integrate

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

Q. Evaluate :

\int_{0}^{1} {sin}^{- 1} (x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}) d x, 0 \leq x \leq 1

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

Join BYJU'S Learning Program

∫120x sin−1x√1−x2dx=

The correct option is B 12−√3π12Put t=sin−1x⇒dt=1√1−x2dx, then ∫120x sin−1x√1−x2dx=∫π60 t sin t dt=[−t cos t+sin t]π60 =[−π6.√32+12]=[12−√3π12]

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