i'm having trouble evaluating the integral at pi/2 and 0.
i know:
s (at pi/2 and 0) sin^2 (2x)dx=
s 1/2[1-cos(2x)]dx=
s 1/2(x-sin(4x))dx=
(x/2)- 1/8[sin (4x)]
i don't understand how you get pi/4
You made a few mistakes, check again. But you don't need to do any calculations to find out that the integral is pi/4. You just note that cos^(2x) integrated over the same interval will yield the same value, because it attains the same values in that interval as sin^(2x) does.
If you add up the two integrals then, because sin^2 + cos^= 1, you get pi/2. SO each integral is pi/4.
To evaluate the integral at π/2 and 0, you can start by using the given expression:
∫[0 to π/2]sin^2(2x)dx = 1/2∫[0 to π/2] (1 - cos(4x)) dx.
Next, you can split the integral into two separate integrals:
1/2(∫[0 to π/2] 1 dx - ∫[0 to π/2] cos(4x) dx).
The first integral, ∫[0 to π/2] 1 dx, simply integrates the constant 1 over the specified interval, resulting in π/2 - 0 = π/2.
The second integral, ∫[0 to π/2] cos(4x) dx, can be evaluated using the formula for integrating cosine:
∫cos(mx) dx = (1/m)sin(mx),
where m is a constant.
Applying the formula, we get:
(1/2)(1/4)sin(4x) | [0 to π/2].
Evaluating the bounds, we get:
(1/8)sin(2π) - (1/8)sin(0).
Since sin(2π) = 0 and sin(0) = 0, these terms cancel out, resulting in a value of 0 for the second integral.
Therefore, the overall value of the integral is:
1/2(π/2) + 0 = π/4.
Hence, the correct value for the integral is π/4, not π/4.