Find the integral from 0 to 2 of xsqrt(5-sqrt4-x^2)).
The hint said to use substitution of u=sqrt(4-x^2), and that I needed one more substitution, but I don't know how to do it.
Sorry I forgot the dx after the integral.
∫[0,2] x√(5-√(4-x^2)) dx
Let
u = 5-√(4-x^2)
du = x/√(4-x^2) dx
so, x dx = (5-u) du
and you now have
∫[3,5] √u (5-u) du
and I'm sure you can handle that, eh?
Wait, why is it the integral from 3-5 instead of 0-2?
because now we are using u, not x.
If, after integration on u, you substitute back into terms of x, then you would use the limits [0,2].
When x=0, u=3 and when x=2, u=5.
To solve the integral ∫[0 to 2] x√(5 - √(4 - x^2)) dx, you are correct in using the substitution u = √(4 - x^2).
This is a standard technique called trigonometric substitution. By substituting u for √(4 - x^2), we can simplify the integral to a more manageable form.
To find the second substitution, we need to consider what other variable we can introduce to simplify the expression further. In this case, we notice that the expression involves a square root of a difference of squares (√(4 - x^2)). Whenever we encounter this square root in trigonometric substitution, it is useful to introduce trigonometric functions such as sine or cosine.
Thus, let's introduce another substitution. Since u = √(4 - x^2) is a positive square root, we can write it as u = 2sinθ, where θ is an angle between -π/2 to π/2 (since sinθ ranges from -1 to 1).
Differentiating both sides with respect to x, we get:
du/dx = 2cosθ * dθ/dx.
Now, let's derive dx in terms of dθ:
Given u = 2sinθ, we rearrange the equation as:
θ = arcsin(u/2).
Differentiating both sides with respect to x, we get:
dθ/dx = (1/√(1 - (u/2)^2)) * (du/dx).
Substituting this into du/dx = 2cosθ * dθ/dx, we can simplify it to:
du/dx = 2cosθ * (1/√(1 - (u/2)^2)) * (du/dx).
Simplifying further, we can cancel out du/dx on both sides:
1 = 2cosθ * (1/√(1 - (u/2)^2)).
Now, let's express x in terms of θ:
Given u = √(4 - x^2) and u = 2sinθ, we can rearrange the equation as:
x = √(4 - u^2) = √(4 - 4sin^2θ) = 2√(1 - sin^2θ) = 2cosθ.
Now, we can modify the original integral using these substitutions:
∫[0 to 2] x√(5 - √(4 - x^2)) dx = ∫[θ1 to θ2] (2cosθ)√(5 - u) * (1/√(1 - (u/2)^2)) du,
where θ1 and θ2 represent the values of θ when x equals 0 and 2, respectively.
Now, we have transformed the original integral in terms of θ and u, making it easier to work with.