Question:
The solution of the equation $\frac{dy}{dx}=\sqrt\frac{1-y^2}{1-x^2}$ is
The solution of the equation $\frac{dy}{dx}=\sqrt\frac{1-y^2}{1-x^2}$ is
Updated On: Jul 7, 2022
- $\sin^{-1}y -\sin^{-1} x = C $
- $ \sin^{-1} y + \sin^{-1} x = C $
- $\sin^{-1} (xy) = 2 $
- None of these
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The Correct Option is A
Solution and Explanation
Given, $\frac{dy}{\sqrt{1-y^{2}}} = \frac{dx}{\sqrt{1-x^{2}}}$
Integrating both sides, we get
$sin^{-1}\, y = sin^{-1}\, x + C$
or $sin^{-1}\, y = sin^{-1}\, x = C$.
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.