Question

Trig interval?

Answer 1

Given function ∫ xsin²x

First find ∫ sin²x dx

Use the Formula : cos(2x) = 1 - 2sin²x ⇒ sin²(x) = ½(1 - cos(2x))

∫ sin²x dx = ∫sin²x dx

∫ sin²x dx = ½ ∫(1 - cos2x) dx

∫ sin²x dx = ½ (x -(sin2x)/2)

∫ sin²x dx = x/2 - (1/4)sin 2x

Now come to Given function ∫ xsin²x

= ∫ xsin²x dx

Apply formula : ∫ UVdx = U∫ Vdx - ∫ [d/dx(U) ∫Vdx] dx

= x ∫ sin²x dx - ∫ [d/dx(x) ∫sin²x dx] dx

Substitute ∫sin²x dx = x/2 - (1/4)sin2x

= x[x/2 - (1/4)sin2x] - ∫[x/2 - (1/4)sin2x] dx

= x²/2 - (x/4)sin2x - (1/2)(x²/2) - (1/4)(cos 2x)/2 + c

= x²/2 - (x/4)sin2x - x²/4 - (1/8)cos2x + c

= x²/4 - (x/4)sin2x - (1/8) cos2x + c

The solution is ∫ xsin²x = x²/4 - (x/4)sin2x - (1/8) cos2x + c