Find the integral of [x^2] * [sin^2(x)] dx?

Answer 1

First note that sin² x = (1-cos(2x))/2, so this is:

1/2 ∫x² - x² cos (2x) dx

We break this into two integrals:

1/2 ∫x² dx - 1/2 ∫x² cos (2x) dx

The first one we integrate speedily:

x³/6 - 1/2 ∫x² cos (2x) dx

To evaluate the second, we proceed by parts. u=x², du=2x dx, v=sin (2x)/2, dv = cos (2x) dx. So this is:

x³/6 - 1/2 (x² sin (2x)/2 - ∫x sin (2x) dx)

Simplifying a bit:

x³/6 - x² sin (2x)/4 + 1/2 ∫x sin (2x) dx

Integrating by parts again: u=x, du=dx, v=-cos (2x)/2, dv=sin (2x) dx

x³/6 - x² sin (2x)/4 + 1/2 (-x cos (2x)/2 + ∫cos (2x)/2 dx)

Simplifying a bit:

x³/6 - x² sin (2x)/4 - x cos (2x)/4 + ∫cos (2x)/4 dx

Computing the last integral:

x³/6 - x² sin (2x)/4 - x cos (2x)/4 + sin (2x)/8 + C

And we are done.

Answer 2

To solve such problem, remember this formula

cos 2x = cos^2(x)-sin^2(x) = 1-2sin^2(x) = 2cos^2(x)-1

x^2*sin^2(x)=x^2*(1/2(1-cos 2x))=1/2*x^2*(1-cos 2x)

integral of x^2 = 1/3*x^3

integral of x^2*cos 2x need to use the integral by part method.
Use int to represent the integral sign

int(x^2 * cos 2x) = ( x^2*(1/2 * sin 2x) - int((2 * x)*(1/2 * sin 2x))
= 1/2 * x^2 * sin 2x - int(x * sin 2x)

int(x * sin 2x) = ( x * -1/2 * cos 2x) - int(1 * (-1/2 * cos 2x))
= -1/2 * x * cos 2x + 1/2 int (cos 2x)
= -1/2 * x * cos 2x + 1/4 sin 2x

integral of [x^2] * [sin^2(x)]
= 1/6*x^3 + 1/4 * x^2 * sin 2x -1/4 * x * cos 2x + 1/8 sin 2x + c

Answer 3

integration by parts.