What is the definite integral from pi divided by 6 to 0 of sin cubed 2x cos cubed 2x dx?

Answer 1

The definite integral from (pi/6) to 0 of:

sin3(2x)cos3(2x) dx

A series of algebraic manipulations using trigonometric identities will be illustrated below:

sin3(2x)cos2(2x)cos(2x) dx

sin3(2x)(1 - sin2(2x))cos(2x) dx

From here, a u-substitution will be used to eliminate the cos(2x) from the expression:

u=sin(2x)

du = 2cos(2x) dx => dx = du/(2cos(x))

making a rewrite of the expression in u:

(1/2)u3(1 - u2) du

(1/2)(u3 - u5) du

From here, elementary antiderivatives show that the indefinite integral from here (in terms of u) is:

(1/2)[(1/4)u4 - (1/6)u6] + C

(1/8)u4 - (1/12)u6 + C

From this point, you could resubstitute back to put the expression back into terms of x using u=sin(2x) and solve using the original limits of integration. Or, more easily, you could put the limits of integration in terms of u:

u = sin(2x)

--- sin(2(pi/6)) = sin(pi/3) = sqrt(3)/2

--- sin(2(0)) = sin(0) = 0

Which turns your problem into:

The integral of (1/2)(u3 - u5) du from sqrt(3)/2 to 0

Using the above antiderivative we already developed, this equals:

(1/8)(sqrt(3)/2)4 - (1/12)(sqrt(3)/2)6 = (1/8)(9/16) - (1/12)(27/64) = (9/128) - (27/768)

So the final answer is whatever the numerical result of this subtraction is:

(9/128) - (27/768)