Integral from pi/6 to pi/4 of 2sin(2x)cos(2x)dx=
2sin(2x)cos(2x)=sin(4x) (double angle)
∫sin(4x) dx = -(1/4)cos(4x) evaluated from π/6 to π/4
=-(1/4)cos(4(π/4)-(-1/4)cos(4(π/6)
=1/4-(-1/4(-1/2)
=1/4 - 1/8
=1/8
To find the integral of 2sin(2x)cos(2x) with respect to x from the lower limit pi/6 to the upper limit pi/4, we can use the regular rules of integration.
The integral of sin(2x)cos(2x) can be simplified using the double angle identity for sine: sin(2θ) = 2sin(θ)cos(θ). In this case, θ = x. So we rewrite the integral as:
2 * 2sin(x)cos(x)cos(2x)dx
Next, we use the double angle identity for cosine: cos(2θ) = cos^2(θ) - sin^2(θ). Here, θ = x. So we rewrite the integral again, substituting cos(2x) with cos^2(x) - sin^2(x):
2 * 2sin(x)(cos^2(x) - sin^2(x))dx
Now, we expand the expression further:
4sin(x)cos^2(x) - 4sin(x)sin^2(x)dx
We can factor out 4 and reorganize the terms:
4(sin(x)cos^2(x) - sin^3(x))dx
Finally, we integrate term by term:
4∫(sin(x)cos^2(x) - sin^3(x))dx
To integrate each term, we can apply the power rule for integration:
For the first term, sin(x)cos^2(x), we can use the substitution u = cos(x). Then du = -sin(x)dx.
Thus, the integral becomes:
4∫(u^2)(-du)
= -4∫u^2 du
= -4 * (u^3 / 3)
= -4u^3 / 3
For the second term, -4∫sin^3(x)dx, we can use the reduction formula:
∫sin^n(x)dx = -(1/n) * sin^(n-1)(x) * cos(x) + (n-1)/n * ∫sin^(n-2)(x)dx
Using this formula, with the reduction step applied twice, we get:
-4 * [-(1/3) * sin^2(x) * cos(x) + (2/3) * ∫sin(x)dx]
= 4/3 * (sin^2(x) * cos(x) - ∫sin(x)dx)
= 4/3 * (sin^2(x) * cos(x) + cos(x))
Now, we substitute back u = cos(x):
= 4/3 * (1 - u^2) * u
= 4/3 * (u - u^3)
= 4/3 * (cos(x) - cos^3(x))
Putting it all together, the integral from pi/6 to pi/4 of 2sin(2x)cos(2x)dx is:
-4u^3 / 3 + 4/3 * (cos(x) - cos^3(x))
Now, let's substitute u = cos(x) back into the expression:
-4(cos(x))^3 / 3 + 4/3 * (cos(x) - (cos(x))^3)
Simplifying further:
-4cos^3(x) / 3 + 4cos(x) / 3 - 4cos^3(x) / 3
= -8cos^3(x) / 3 + 4cos(x) / 3
Therefore, the value of the integral from pi/6 to pi/4 of 2sin(2x)cos(2x)dx is -8cos^3(x) / 3 + 4cos(x) / 3.