-01-2sin-1x-1-x23-2dx-

I = ∫_0^1/√(2)sin^ 1x/(1 x^2)^3/2dx Put sin^ 1x=t ⇒1/√(1 x^2)dx=dt and x=sin t Also t=0 to π/4as x=0 to 1/√(2) ⇒ I=∫_0^π/4 t.sec^2 t dt=π/4 1/2log 2 ...

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Byju's Answer

Standard XI

Mathematics

Property 1

∫01/√2sin-1x/...

Question

$\int_{0}^{\frac{1}{\sqrt{2}}} \frac{s i n^{- 1} x}{(1 - x^{2})^{\frac{3}{2}}} d x =$

\frac{π}{4} + \frac{1}{2} l o g 2

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\frac{π}{4} - \frac{1}{2} l o g 2

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\frac{π}{2} + l o g 2

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\frac{π}{2} - l o g 2

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Solution

The correct option is B $\frac{π}{4} - \frac{1}{2} l o g 2$
$I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{s i n^{- 1} x}{(1 - x^{2})^{\frac{3}{2}}} d x$
$P u t s i n^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t a n d x = s i n t$
$A l s o t = 0 t o \frac{π}{4} a s x = 0 t o \frac{1}{\sqrt{2}}$
$\Rightarrow I = \int_{0}^{\frac{π}{4}} t . s e c^{2} t d t = \frac{π}{4} - \frac{1}{2} l o g 2$

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∫1√20sin−1x(1−x2)32dx=

The correct option is B π4−12log 2I=∫1√20sin−1x(1−x2)32dx Put sin−1x=t⇒1√1−x2dx=dt and x=sin t Also t=0 to π4as x=0 to 1√2 ⇒I=∫π40 t.sec2 t dt=π4−12log 2

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