How would you solve the following problem explicitly?
Sqrt(1-y^2) dx - sqrt(1-x^2) dy
I separated the x and y terms and got:
Integral of 1/sqrt(1-x^2) dx = Integral of 1/sqrt(1-y^2) dy
I was wondering how would you take the anti-derivative of each function. I believe we are using a trig substitution?
Oops, Sqrt(1-y^2) dx - sqrt(1-x^2) dy = 0
sqrt(1-y^2) dx - sqrt(1-x^2) dy = 0
sqrt(1-y^2) dx = sqrt(1-x^2) dy
Take all term swith x to one side and all terms of y to the other side of equation:
dx / sqrt(1-x^2) = dy / sqrt(1-y^2)
Integrate both side. Note that,
∫(1/(sqrt(a^2 - x^2))dx = sin^-1 (x/a) + C
Therefore,
sin^-1 (x) + C = sin^-1 (y)
Get the sine of both sides:
y = sin( sin^-1 (x) + C )
hope this helps? `u`
If we pick a different c, the solution looks a little more symmetric, and more like the original equation:
y = sin(arcsin(x)+arcsin(c))
y = sin(arcsin(x))cos(arcsin(c)) + cos(arcsin(x))sin(arcsin(c))
= x√(1-c^2) + c√(1-x^2)
Just a thought. A bit ugly because c appears twice, and is more restricted than the general C above.
To solve the problem explicitly, you have correctly separated the variables and obtained the equation:
Integral of 1/sqrt(1-x^2) dx = Integral of 1/sqrt(1-y^2) dy
To evaluate these integrals, we can indeed use trigonometric substitutions. Let's start by considering the integral on the left-hand side:
∫ 1/√(1-x^2) dx
To introduce a trigonometric substitution, we can let x = sinθ, which implies dx = cosθ dθ. Substituting these values into the integral, we get:
∫ 1/√(1-sin^2θ) cosθ dθ
Simplifying further, we have:
∫ cosθ/√(cos^2θ) dθ
Since cos^2θ = 1 - sin^2θ, we can rewrite the integral as:
∫ cosθ/√(1-sin^2θ) dθ
Using the Pythagorean identity sin^2θ + cos^2θ = 1, we have:
∫ cosθ/√(cos^2θ) dθ
This simplifies to:
∫ dθ
The integral of 1 with respect to θ is simply θ + C, where C is the constant of integration.
So, for the left-hand side, the anti-derivative is:
θ + C
Now, let's consider the right-hand side integral:
∫ 1/√(1-y^2) dy
Using the same approach as before, we can let y = sinφ, which implies dy = cosφ dφ. Substituting these into the integral, we get:
∫ 1/√(1-sin^2φ) cosφ dφ
Simplifying further, we have:
∫ cosφ/√(cos^2φ) dφ
Using the same trigonometric identity as before, we can rewrite the integral as:
∫ dφ
The integral of 1 with respect to φ is simply φ + C, where C is the constant of integration.
So, for the right-hand side, the anti-derivative is:
φ + C
Now, we have:
θ + C = φ + C
Since the constants of integration cancel out, we can conclude that θ = φ. This means that x = sinθ = sinφ and y = sinφ as well.
Therefore, the solution to the original problem is x = y = sinθ.
I hope this explanation helps you understand how to tackle the trigonometric substitution and take the antiderivative of each function.