I have two questions, because I'm preparing for a math test on monday.
1. Use the fundamental theorem of calculus to find the derivative:
(d/dt) the integral over [0, cos t] of (3/5-(u^2))du
I have a feeling I will be able to find the derivative easily, I'm just having trouble with the very first step-- finding the integral. The only thought I've had so far is possibly rearranging the function like this:
3(5-(u^2))^-1
but I don't know if can work like that. Any ideas?
2. Evaluate these trigonometric integrals:
integral of sin(^2)x*cosx(dx)
I think I may have to use a substitution here, but I'm not sure. I started like this:
(sin(x))^2 + (cos(x))^2 = 1
therefore (sin(x))^2=1-(cos(x))^2
so:
the integral of (1-(cos(x))^2)cosx dx
which brings us to:
integral of cosx-(cos(x))^3dx
Now I know I could split that up into two integrals, but we haven't learned the integral of (cos(X))^3. I could make a substitution where u=cosx and du/-sinx= dx, but then my integral would be:
integral of (u^3)(-cscx)du
which doesn't make the problem any more simple. Any ideas? Was it wrong to make the substitution for (sin(x))^2 at the start?
1. To find the derivative of the given integral, we can apply the fundamental theorem of calculus. This theorem states that if a function F(x) is the integral of another function f(x), then the derivative of F(x) with respect to x is equal to f(x).
In this case, the function to be integrated is (3/5-(u^2))du. To find the integral, you correctly rearranged the function as 3/(5-(u^2)) (assuming there are no errors in your original expression). In order to evaluate this integral, you can use trigonometric substitution.
Let's substitute u = sin(θ) to simplify the integrand. Since u = sin(θ), du = cos(θ)dθ. Also, when u = 0, θ = 0, and when u = cos(t), θ = π/2 - t.
Now the integral becomes:
∫[0, cos(t)] 3/(5 - (sin(θ))^2) * cos(θ) dθ
Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we have:
∫[0, cos(t)] 3/(5 - (1 - (cos(θ))^2)) * cos(θ) dθ
Simplifying further:
∫[0, cos(t)] 3/(4 + (cos(θ))^2) * cos(θ) dθ
To evaluate this integral, you can make the substitution z = sin(θ) and calculate its antiderivative using standard techniques like partial fractions or inverse trigonometric functions. Once you have the antiderivative, you can differentiate it with respect to t to find the derivative of the original integral.
2. For the second question, the integral is:
∫ sin^2(x) * cos(x) dx
You correctly expressed sin^2(x) as 1 - cos^2(x), leading to:
∫ (1 - cos^2(x)) * cos(x) dx
Expanding this expression:
∫ (cos(x) - cos^3(x)) dx
The integral of cos(x) with respect to x is sin(x). However, you are unsure of how to proceed with the integral of cos^3(x) since you haven't learned it yet.
In this case, you can use integration by parts. Let's denote:
u = cos^2(x) and dv = cos(x) dx
By differentiating u, you get:
du = -2cos(x)sin(x) dx
By integrating dv, you get:
v = sin(x)
Applying the integration by parts formula:
∫(cos^3(x)) dx = u*v - ∫v*du
∫(cos^3(x)) dx = cos^2(x)*sin(x) - ∫(-2cos(x)sin(x) * sin(x) dx)
∫(cos^3(x)) dx = cos^2(x)*sin(x) + 2∫cos(x)sin^2(x) dx
The integral 2∫cos(x)sin^2(x) dx can be evaluated using a substitution. Let's substitute u = sin(x), then du = cos(x) dx:
2∫cos(x)sin^2(x) dx = 2∫u^2 du = 2(u^3)/3 + C = 2(sin^3(x))/3 + C
Finally, substituting these results back into the original integral:
∫ sin^2(x) * cos(x) dx = sin(x) - (cos^2(x)*sin(x) + 2(sin^3(x))/3) + C
= sin(x) - cos^2(x)*sin(x) - 2(sin^3(x))/3 + C
So, that is the value of the integral.