integral[13,4]sin^2(8x)cos^2(8x)dx
sin^2(8x)cos^2(8x)
= 1/4 (2sin(8x)cos(8x))^2
= 1/4 sin^2(16x)
now use integration by parts twice to evaluate.
or, even easier, recall your half-angle formula to show that this is
1/2 (1-cos(32x))
If i use integration by parts can i take out 1/4 from the integral and make u=sin^2(16x) and dv=dx ?
To find the integral of sin^2(8x)cos^2(8x)dx between the limits of 13 and 4, you can use two trigonometric identities:
1. sin^2(2θ) = (1 - cos(4θ)) / 2
2. cos^2(2θ) = (1 + cos(4θ)) / 2
Using these identities, we can rewrite sin^2(8x)cos^2(8x) as:
sin^2(8x)cos^2(8x) = (sin^2(2*8x))(cos^2(2*8x))
= ((1 - cos(4*8x)) / 2)((1 + cos(4*8x)) / 2)
Now, the integral can be expressed as:
∫[4,13] (1 - cos(4*8x))(1 + cos(4*8x)) / 4 dx
Expanding and simplifying the expression inside the integral gives:
∫[4,13] (1 - cos^2(4*8x)) / 4 dx
= ∫[4,13] (1 - cos^2(32x)) / 4 dx
Next, we can apply another trigonometric identity: cos^2(θ) = (1 + cos(2θ)) / 2
Substituting this identity into the integral, we get:
∫[4,13] (1 - (1 + cos(2*32x)) / 2) / 4 dx
= ∫[4,13] (1 - (1 + cos(64x)) / 2) / 4 dx
= ∫[4,13] (1 - 1/2 - cos(64x)/2) / 4 dx
= ∫[4,13] (1/2 - cos(64x)/2) / 4 dx
Now, we can separate the terms and integrate each one:
= (1/2) ∫[4,13] dx/4 - (1/2) ∫[4,13] cos(64x)/2 dx
= 1/8 ∫[4,13] dx - 1/4 ∫[4,13] cos(64x) dx
The integral of dx is just x, and the integral of cos(64x) can be found using the substitution u = 64x:
= 1/8 (x) ∣[4,13] - 1/4 ∫[4,13] cos(u) du/64
= 1/8 (x) ∣[4,13] - 1/256 ∫[4,13] cos(u) du
Integrating cos(u) gives sin(u), so the integral becomes:
= 1/8 (x) ∣[4,13] - 1/256 (sin(u)) ∣[4,13]
= 1/8 (x) ∣[4,13] - 1/256 (sin(64x)) ∣[4,13]
Now, substitute the limits of integration back in:
= 1/8 (13) - 1/256 (sin(64*13)) - (1/8) (4) + 1/256 (sin(64*4))
= 13/8 - 1/256 (sin(832)) - 1/2 + 1/256 (sin(256))
Therefore, the value of the integral of sin^2(8x)cos^2(8x)dx between 13 and 4 is 13/8 - 1/256 (sin(832)) - 1/2 + 1/256 (sin(256)).