Hello! I have this problem:
x(dx)/sqrt(9-x^2)
I was wondering why I can't use trig substitution and substitute sqrt(9-x^2) for sqrt(1-sec^2) and having:
integral x = 3sin(theta)
dx = 3cos(theta)d(theata)
integral 3sin(theta)(3cos(theta))/3cos(theta)
having the 3cos(theta) cancel and leaving integral 3sin(theta) d(theta)
equalling 3cos(theta) and coming to an answer of sqrt(9-x^2)/3 + C?
Thank You!
you can do the trig stuff, but why bother?
let u = 9-x^2
du = -2x dx
and the integrand becomes
-1/2 u^(-1/2) du
and integrates to
- u^(1/2) = -√(9-x^2)
As for the trig, you cannot take √(1-sec^2) since sec^2 is always greater than 1, and you don't have a real value.
I'd so
x = 3sinθ
dx = 3cosθ dθ
√(9-x^2) = √(9-9sin^2θ) = 3cosθ
x/√(9-x^2) dx = 3sinθ/3cosθ * 3cosθ dθ
= 3sinθ dθ
the integral is then just
-3cosθ = -√(9-x^2) + C
You lost a factor of 3 in there somewhere.
Hello! I understand that you are trying to integrate the expression x(dx)/sqrt(9-x^2) and you attempted to use trigonometric substitution by substituting sqrt(9-x^2) for sqrt(1-sec^2). However, there is a mistake in your calculation.
Let me explain the correct way to solve this integral using trigonometric substitution:
To begin, let's make the substitution x = 3sin(theta). This substitution is valid because it maps the interval [-3, 3] to the interval [-pi/2, pi/2], which is the range of theta for the integral.
Differentiating both sides with respect to theta, we get dx = 3cos(theta)d(theta).
Now let's substitute these expressions into the integral:
∫(x(dx))/sqrt(9-x^2) = ∫((3sin(theta))(3cos(theta)))/(sqrt(9-(3sin(theta))^2))d(theta)
Simplifying the numerator, we have 9sin(theta)cos(theta). And for the denominator, we can apply the Pythagorean identity: 9 - (3sin(theta))^2 = 9 - 9sin^2(theta) = 9cos^2(theta).
So the integral becomes:
∫(9sin(theta)cos(theta))/(sqrt(9cos^2(theta)))d(theta)
The square root and the squared cosine cancel out, leaving:
∫9sin(theta)d(theta)
Integrating 9sin(theta) with respect to theta gives us -9cos(theta) + C, where C is the constant of integration.
Now, let's substitute back x = 3sin(theta) to convert our answer back to the original variable:
-9cos(theta) + C = -9cos(arcsin(x/3)) + C
Using the identity cos(arcsin(u)) = sqrt(1 - u^2), we can simplify further:
-9cos(arcsin(x/3)) + C = -9sqrt(1 - (x/3)^2) + C
Therefore, the correct answer should be -9sqrt(1 - (x/3)^2) + C.
I hope this clarifies why your previous approach using trigonometric substitution did not yield the correct result. If you have any further questions, feel free to ask!