If - x -lim n -x n - x - n - x n - x - n - 0- x -1- - N- then -sin -1 x - x dx is equal toA- None of these B- x sin -1 x -1- x 2- CC- - x sin -1 x -1- x 2D- x sin -1 x-1-x2-C

The correct option is B x sin^ 1(x)+√(1 x^2)+CWe have ϕ (x) = lim_n→∞x^2n 1/x^2n+1=1 ,0 lt;x lt;1 ∴∫ sin^ 1x.ϕ(x) dx=∫ sin^ 1x dx =[x sin^ 1 x + √(1 x^2 ...

If Φ x =lim n...

Question

If $Φ (x) = {lim}_{n \to \infty} \frac{x^{n} - x^{- n}}{x^{n} + x^{- n}}, 0 < x < 1, ϵ N$ , then $\int (s i n^{- 1} x) (Φ (x)) d x$ is equal to

x s i n^{- 1} (x) + \sqrt{1 - x^{2}} + C

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- (x s i n^{- 1} (x) + \sqrt{1 - x^{2}})

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x s i n^{- 1 x + \sqrt{1 - x^{2}} + C}

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None of these

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Solution

The correct option is B $x s i n^{- 1} (x) + \sqrt{1 - x^{2}} + C$
We have $ϕ (x) = {lim}_{n \to \infty} \frac{x^{2 n} - 1}{x^{2 n} + 1} = 1, 0 < x < 1 ∴ \int s i n^{- 1} x . ϕ (x) d x = \int s i n^{- 1} x d x = [x s i n^{- 1} x + \sqrt{1 - x^{2}}] + c$

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The correct option is B x sin−1(x)+√1−x2+CWe have ϕ(x)=limn→∞x2n−1x2n+1=1,0<x<1∴∫sin−1x.ϕ(x) dx=∫sin−1x dx=[x sin−1 x+√1−x2]+c

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The correct option is B $x s i n^{- 1} (x) + \sqrt{1 - x^{2}} + C$
We have $ϕ (x) = {lim}_{n \to \infty} \frac{x^{2 n} - 1}{x^{2 n} + 1} = 1, 0 < x < 1 ∴ \int s i n^{- 1} x . ϕ (x) d x = \int s i n^{- 1} x d x = [x s i n^{- 1} x + \sqrt{1 - x^{2}}] + c$