integral of x^2 ( 1 - 2sin^2 (x) ) dx

Question

i need help

Answer 1

1-2sin²(x) can be replaced by cos(2x), so the integrand is x²cos(2x).

Let u=x², then du/dx=2x; let dv/dx=cos(2x), then v=½sin(2x).

Integrating by parts: uv-∫vdu=½x²sin(2x)-∫xsin(2x)dx.

Let u=x, then du/dx=1; let dv/dx=sin(2x), then v=-½cos(2x).

Integrating by parts: ∫xsin(2x)dx=-½xcos(2x)+½∫cos(2x)dx=

-½xcos(2x)+¼sin(2x)dx, therefore, ∫x²cos(2x)dx=½x²sin(2x)+½xcos(2x)-¼sin(2x)+C, where C is the integration constant.

CHECK

Differentiate ½x²sin(2x)+½xcos(2x)-¼sin(2x)+C:

x²cos(2x)+xsin(2x)-xsin(2x)+½cos(2x)-½cos(2x)=x²cos(2x)=x²(1-2sin²(x)).