Let f(x)=cosxsqrt(1+sinx)
A. If F(x)=the integral of f(x)dx and F(0)=5/3, find F(pi/2).
B. If G(x)=the integral of f(x)sinx dx and G(0)=1/3, find G(x).
Try using the half-angle formulas to re-express the integrands.
I don't understand your explanation. How would you go about it with u-substitution?
To find the value of F(pi/2), we need to evaluate the definite integral of f(x) with respect to x from 0 to pi/2.
A. First, let's find the antiderivative of f(x) to get F(x):
F(x) = ∫[0, x] f(t) dt
To find F(x), we need to find the indefinite integral of f(x) with respect to x. Therefore, we integrate f(x) term by term.
Let's start with each term:
∫ cos(x) dx = sin(x) + C1 (where C1 is a constant of integration)
∫ sqrt(1 + sin(x)) dx:
This integral requires a substitution:
Let u = 1 + sin(x)
Then, du = cos(x) dx
∫ sqrt(u) du:
= (2/3) u^(3/2) + C2 (where C2 is a constant of integration)
Therefore, we have:
F(x) = sin(x) + (2/3) (1 + sin(x))^(3/2) + C (where C = C1 + C2)
Given that F(0) = 5/3, we can substitute this value into the equation:
5/3 = sin(0) + (2/3) (1 + sin(0))^(3/2) + C
5/3 = 0 + (2/3) (1 + 0)^(3/2) + C
5/3 = 2/3 + C
C = 5/3 - 2/3
C = 3/3
C = 1
So, F(x) = sin(x) + (2/3) (1 + sin(x))^(3/2) + 1
Now we can evaluate F(pi/2):
F(pi/2) = sin(pi/2) + (2/3) (1 + sin(pi/2))^(3/2) + 1
F(pi/2) = 1 + (2/3) (1 + 1)^(3/2) + 1
F(pi/2) = 1 + (2/3) (2)^(3/2) + 1
F(pi/2) = 1 + (2/3) (2 * sqrt(2)) + 1
F(pi/2) = 1 + (4/3) sqrt(2) + 1
F(pi/2) = 2 + (4/3) sqrt(2)
Therefore, F(pi/2) = 2 + (4/3) sqrt(2)
B. To find G(x), we need to evaluate the definite integral of f(x)sin(x) with respect to x from 0 to x and use the initial condition G(0) = 1/3.
G(x) = ∫[0, x] f(t)sin(t) dt
Let's start with each term:
∫ cos(t)sin(t) dt:
This integral can be evaluated by using a substitution:
Let u = cos(t), then du = -sin(t) dt
The integral becomes:
-∫ u du
= -u^2/2 + C3 (where C3 is a constant of integration)
∫ sqrt(1 + sin(t))sin(t) dt:
This integral can be evaluated by using a substitution as well:
Let u = 1 + sin(t), then du = cos(t) dt
The integral becomes:
∫ sqrt(u) du
= (2/3) u^(3/2) + C4 (where C4 is a constant of integration)
Therefore, we have:
G(x) = -cos(x)^2/2 + (2/3) (1 + sin(x))^(3/2) + C5 (where C5 = C3 + C4)
Given that G(0) = 1/3, we can substitute this value into the equation:
1/3 = -cos(0)^2/2 + (2/3) (1 + sin(0))^(3/2) + C5
1/3 = -1/2 + (2/3) (1 + 0)^(3/2) + C5
1/3 = -1/2 + (2/3) (1)^(3/2) + C5
1/3 = -1/2 + 2/3 + C5
C5 = 1/3 - 2/3 + 1/2
C5 = -1/3 + 1/2
C5 = 3/6 - 2/6
C5 = 1/6
So, G(x) = -cos(x)^2/2 + (2/3) (1 + sin(x))^(3/2) + 1/6
Therefore, G(x) = -cos(x)^2/2 + (2/3) (1 + sin(x))^(3/2) + 1/6.