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- x sin -1 x -1- x 2- CB- - x sin -1 x -1- x 2C- x sin -1 x-1-x2-CD- None of these

The correct option is A 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 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

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

Q. 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

Q. 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

Q. If

y = x \sin^{- 1} x + \sqrt{1 - x^{2}}

, prove that

\frac{d y}{d x} = \sin^{- 1} x

Q. If

y = \frac{x {sin}^{- 1} x}{\sqrt{1 - x^{2}}},

prove that

(1 - x^{2}) \frac{d y}{d x} = x + \frac{y}{x}

\int {[x s i n^{- 1} x {\sqrt{(1 - x^{2}}}^{- 1}]}

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If Φ(x)=limn→∞xn−x−nxn+x−n,0<x<1,ϵN, then ∫(sin−1x)(Φ(x))dx is equal to

The correct option is A 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

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

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