Integrate((1-Cos((Pi*x)/(2*a)))^2,x,0,a)
Question
Integrate((1-Cos((Pix)/(2a)))^2,x,0,a)
Solution
To solve the integral ∫((1-Cos((Pix)/(2a)))^2 dx from 0 to a, we can follow these steps:
-
First, expand the square in the integrand: (1 - Cos((Pix)/(2a)))^2 = 1 - 2Cos((Pix)/(2a)) + Cos^2((Pix)/(2a))
-
Now, we can split the integral into three separate integrals: ∫dx - 2∫Cos((Pix)/(2a)) dx + ∫Cos^2((Pix)/(2a)) dx from 0 to a.
-
The first integral is straightforward: ∫dx from 0 to a = [x]_0^a = a.
-
The second integral can be solved using the formula ∫Cos(u) du = Sin(u) + C. Here, u = (Pix)/(2a), so du = (Pi/2a) dx. Therefore, ∫Cos((Pix)/(2a)) dx = 2aSin((Pix)/(2a)) from 0 to a = 2a[0 - Sin(Pi/2)] = -2a.
-
The third integral, ∫Cos^2((Pix)/(2a)) dx from 0 to a, can be solved using the power-reduction identity: Cos^2(u) = 1/2 + 1/2Cos(2u). Therefore, ∫Cos^2((Pix)/(2a)) dx = 1/2∫dx + 1/2∫Cos(Pi*x/a) dx from 0 to a. The first integral is a/2, and the second integral is 0 (because the integral of Cos over a full period is 0).
-
Therefore, the original integral is a - 2a + a/2 = a/2.
Similar Questions
integrate(x^3*cos x/2+1/2)*sqrt(4-x^2)dx from-2 to 2
Find by double integration the area of the region enclosed by curves2 2 2 ,x y a x y a in the first quadrant.
Show that d/dalpha (integrate arctan(x/2) dx from 0 to a ^ 2) = 2a * arctan(alpha) - 1/2 * log(alpha ^ 2 + 1)
Evaluate the integral. (Use C for the constant of integration.)(x2 + 2x) cos(x) dx
Evaluate the following integral:∫0𝜋/2𝑥sin(𝑥) 𝑑𝑥∫ 0π/2 xsin(x)dx
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.