Question 1
�ç e^2x Cos(4x) dx
Question 2)
�ç �ãx sin(x) dx
Question 3)
�ç x^2 sin(x) dx
I'll do #1, you can show us where you get stuck on the others.
Let I = ∫e^2x cos4x dx
u=cos 4x
du = -4sin4x dx
dv = e^2x dx
v = 1/2 e^2x
I = uv - ∫v du
= 1/2 e^2x cos4x - ∫-2e^2x sin4x dx
Now do it again, letting u = sin4x
I = 1/2 e^2x cos4x + (e^2x sin4x - ∫4e^2x cos4x dx)
= 1/2 e^2x cos4x + e^2x sin4x - 4I
5I = 1/2 e^2x cos4x + e^2x sin4x
I = 1/10 e^2x (cos4x + 2sin4x)
To solve these integration problems, we can use integration by parts, which is a technique based on the product rule of differentiation. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
where u and v are functions of x.
Let's solve each question step by step.
Question 1:
∫ e^(2x) * cos(4x) dx
To use integration by parts, we need to choose which part to differentiate and which part to integrate. Let's assign:
u = e^(2x) (to be differentiated)
dv = cos(4x) dx (to be integrated)
Now, let's find du and v:
du = d/dx(e^(2x)) dx = 2e^(2x) dx
v = ∫ cos(4x) dx = (1/4) sin(4x)
Applying the formula:
∫ e^(2x) * cos(4x) dx = (e^(2x) * (1/4) sin(4x)) - ∫ 2e^(2x) * (1/4) sin(4x) dx
Simplifying:
∫ e^(2x) * cos(4x) dx = (1/4) e^(2x) sin(4x) - (1/2) ∫ e^(2x) sin(4x) dx
Now, we have a new integral in the right term:
∫ e^(2x) sin(4x) dx
To solve it, we can again use integration by parts. Assign:
u = sin(4x) (to be differentiated)
dv = e^(2x) dx (to be integrated)
Compute du and v:
du = d/dx(sin(4x)) dx = 4 cos(4x) dx
v = ∫ e^(2x) dx = (1/2) e^(2x)
Apply the formula again:
∫ e^(2x) sin(4x) dx = (1/2) e^(2x) sin(4x) - ∫ (1/2) 4e^(2x) cos(4x) dx
= (1/2) e^(2x) sin(4x) - 2 ∫ e^(2x) cos(4x) dx
But we already know the value of the integral on the right side from before:
∫ e^(2x) cos(4x) dx = (1/4) e^(2x) sin(4x) - (1/2) ∫ e^(2x) sin(4x) dx
Now we can substitute it back into the equation:
∫ e^(2x) sin(4x) dx = (1/2) e^(2x) sin(4x) - 2 * ((1/4) e^(2x) sin(4x) - (1/2) ∫ e^(2x) sin(4x) dx)
Now, rearrange the equation and isolate the integral term:
∫ e^(2x) sin(4x) dx = (1/2) e^(2x) sin(4x) - (1/2) e^(2x) sin(4x) + ∫ e^(2x) sin(4x) dx
Simplifying:
∫ e^(2x) sin(4x) dx = e^(2x) sin(4x)
So, the answer to the first question is:
∫ e^(2x) cos(4x) dx = (1/4) e^(2x) sin(4x) - (1/2) ∫ e^(2x) sin(4x) dx
= (1/4) e^(2x) sin(4x) - (1/2) (e^(2x) sin(4x))
= (1/4) e^(2x) sin(4x) - (1/2) e^(2x) sin(4x)
= (1/4 - 1/2) e^(2x) sin(4x)
= -(1/4) e^(2x) sin(4x)
Now let's move on to the second question:
Question 2:
∫ x * sin(x) dx
For this integral, we can use integration by parts. Assign:
u = x (to be differentiated)
dv = sin(x) dx (to be integrated)
Find du and v:
du = d/dx(x) dx = dx
v = ∫ sin(x) dx = -cos(x)
Apply the formula:
∫ x * sin(x) dx = -x * cos(x) + ∫ cos(x) dx
= -x * cos(x) + sin(x)
So the answer to the second question is:
∫ x * sin(x) dx = -x * cos(x) + sin(x)
Finally, let's solve the third question:
Question 3:
∫ x^2 * sin(x) dx
Again, we can use integration by parts. Assign:
u = x^2 (to be differentiated)
dv = sin(x) dx (to be integrated)
Find du and v:
du = d/dx(x^2) dx = 2x dx
v = ∫ sin(x) dx = -cos(x)
Apply the formula:
∫ x^2 * sin(x) dx = -x^2 * cos(x) + ∫ 2x * cos(x) dx
= -x^2 * cos(x) + 2 ∫ x * cos(x) dx
We already know how to solve the integral on the right side from the second question:
∫ x * sin(x) dx = -x * cos(x) + sin(x)
Substitute it back into the equation:
∫ x^2 * sin(x) dx = -x^2 * cos(x) + 2 * (-x * cos(x) + sin(x))
= -x^2 * cos(x) - 2x * cos(x) + 2sin(x)
Therefore, the answer to the third question is:
∫ x^2 * sin(x) dx = -x^2 * cos(x) - 2x * cos(x) + 2sin(x)