Q.15. Integration

Question 15.
Find ∫x^2 * sin^2 x dx

Answer 15.
∫x^2*sin^2 x dx
= (1/2) ∫x^2 * 2sin^2 x dx
= (1/2) ∫x^2 * (1 - cos2x) dx
= (1/2) ∫x^2 dx - (1/2) ∫x^2 * cos2x dx
= x^3/6 - (1/2) ∫x^2 * cos2x dx ... (1)

∫x^2 * cos2x dx
Using the formula for integration by parts,
∫uv dx = u∫vdx - ∫[du/dx∫vdx] dx,
Taking u = x^2 and v = cos2x,
Integral
= x^2 ∫cos2x dx - ∫[d/dx(x^2)∫cos2x dx] dx
= (1/2)x^2 * sin2x - ∫[(2x) * (1/2) sin2x] dx
= (x^2/2) sin2x - ∫x sin2x dx
Using integration by parts again with u = x and v = sin2x,
= (x^2/2) sin2x - [x∫sin2x dx - ∫[d/dx(x)∫sin2x dx] dx]]
= (x^2/2) sin2x - x∫sin2x dx + ∫-(1/2)cos2x dx
= (x^2/2) sin2x + (x/2)cos2x - (1/4)sin2x - 2c ... (2)

From (1) and (2),
∫x^2*sin^2 x dx
= x^3/6 - (1/2)[(x^2/2) sin2x + (x/2)cos2x - (1/4)sin2x - 2c]
= x^3/6 - (1/8) (2x^2 - 1) sin2x - (x/4) cos2x + c.

Link to YA! 
[ I found that this question has been deleted from YA! However, I have decided to retain it in my blog. ]