Integral of XarcsinX dx. ii)X^2.arcsinX dx

Answer 1

do these the same way, using integration by parts

∫x arcsin(x) dx
Let arcsin x, du = 1/√(1-x^2) dx
dv = x dx, v = 1/2 x^2

∫ u dv = uv - ∫ v du
∫ x arcsin(x) dx = 1/2 x^2 arcsin(x) - ∫x^2 / 2√(1-x^2) dx

Now you should be able to proceed.

Answer 2

I got stuck here,can you please proceed/complete it

Answer 3

now let

x = sinθ
dx = cosθ dθ

∫x^2 / 2√(1-x^2) dx
= ∫ sin^2θ / 2cosθ * cosθ dθ
= 1/2 ∫sin^2θ dθ
= 1/4 ∫1-cos2θ dθ
= 1/4 (θ - 1/2 sin 2θ)
= 1/4 (θ - sinθ cosθ)
= 1/4 (arcsin(x) - x√(1-x^2))

Now just put it all together

Answer 4

It occurs to me that a simpler way to go is

x = sinθ, θ = arcsin(x)
dx = cosθ dθ

∫x arcsin(x) dx = ∫ θ sinθ cosθ dθ
= 1/2 ∫ θ sin2θ dθ

Now use integration by parts to get
u = θ, du = dθ
dv = sin2θ dθ, v = -1/2 cos2θ

and there's a lot less messing around.