I am looking for help integrating using u substitution over the interval of 0 to pi/2 the function cos(2x/3)
the perimeter of parallelogram is 52cm,and jk is 2cm longer than jm.
find jk
find jm
Integrate cos (2x/3) dx
u = 2x/3
du = 2/3 dx
3/2 du = dx
I = integrate symbol
I cos (2x/3) dx = I cos u 3/2 du
I cos (2x/3) dx = 3/2 I cos u du
I cos (2x/3) dx = 3/2 sins u
since, u = 2x/3
I cos (2x/3) dx = 3/2 sin (2x/3)
I cos (2x/3) dx = 3/2 sin (2x/3)
sorry, forgot 0 to pi/2
3/2 sin 2/3 (pi/2) - 3/2 sin 0
3/2 sin (2pi/6) - 0
3/2 sin (pi/3)
3/2 (sqrt3/2)
3 sqrt3/4
To integrate the function cos(2x/3) using u-substitution, follow these steps:
Step 1: Choose a suitable substitution
In this case, we can let u = 2x/3. This choice simplifies the function and makes the integration easier.
Step 2: Calculate du
Differentiate u with respect to x to find du. Since u = 2x/3, we have du/dx = 2/3. Rearranging, we get du = (2/3)dx.
Step 3: Substitute u and du into the integral
Substitute the expression for u and du into the original integral:
∫ cos(2x/3) dx
Substituting u = 2x/3, we get 3/2 * du = dx. Replacing dx with 3/2 * du, the integral becomes:
∫ cos(u) * (3/2) du
Step 4: Evaluate the integral
We can now integrate with respect to u, treating cos(u) as a constant:
(3/2) ∫ cos(u) du = (3/2) * sin(u) + C
where C represents the constant of integration.
Step 5: Replace u in terms of x
To obtain the final answer in terms of x, substitute u = 2x/3 back into the expression:
(3/2) * sin(u) + C = (3/2) * sin(2x/3) + C
Thus, the integration of cos(2x/3) over the interval 0 to π/2 is:
∫[0,π/2] cos(2x/3) dx = (3/2) * sin(2π/6) + C - (3/2) * sin(0) + C
Simplifying further:
= (3/2) * sin(π/3) - (3/2) * 0
= (3/2) * (√3/2)
= (3√3) / 4