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

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.

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

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

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.

To find the integral of Xarcsin(X) dx, we can use integration by parts. Here are the steps:

Step 1: Identify u and dv
Let u = X and dv = arcsin(X) dx

Step 2: Find du and v
To find du, we differentiate u with respect to X.
du = dX

To find v, we integrate dv with respect to X.
∫dv = ∫arcsin(X) dx

This can be solved by using integration by substitution. Let's say X = sin(t), then dX = cos(t) dt. The integral becomes:
∫arcsin(X) dx = ∫t cos(t) dt

Step 3: Apply the integration by parts formula
The integration by parts formula states:
∫u dv = uv - ∫v du

Using this formula, we have:
∫Xarcsin(X) dx = X ∫arcsin(X) dx - ∫(∫arcsin(X) dx) dX

Step 4: Solve the remaining integrals
∫arcsin(X) dx = ∫t cos(t) dt

We can integrate this by using integration by parts again.
Let u = t and dv = cos(t) dt.

du = dt
∫dv = ∫cos(t) dt = sin(t)

By applying the integration by parts formula, we get:
∫t cos(t) dt = t sin(t) - ∫sin(t) dt

The integral of sin(t) can be easily found:
∫sin(t) dt = -cos(t)

Therefore:
∫t cos(t) dt = t sin(t) - (-cos(t))

Simplifying further:
∫t cos(t) dt = t sin(t) + cos(t)

Step 5: Final result
Substituting back into the equation from step 3:
∫Xarcsin(X) dx = X(t sin(t) + cos(t)) - ∫(t sin(t) + cos(t)) dX

Since X = sin(t),
∫Xarcsin(X) dx = sin(t)(t sin(t) + cos(t)) - ∫(sin(t)(t sin(t) + cos(t))) dX

Therefore, the final answer is:
∫Xarcsin(X) dx = X(sin(t)(t sin(t) + cos(t))) - ∫(sin(t)(t sin(t) + cos(t))) dX

Similarly, to find the integral of X^2.arcsin(X) dx, you can follow the same steps but with slight modifications.

Step 1: Identify u and dv
Let u = X^2 and dv = arcsin(X) dx

Step 2: Find du and v
Differentiate u with respect to X to find du.
du = 2X dx

Integrate dv with respect to X to find v.
∫dv = ∫arcsin(X) dx

Again, using u-substitution, let's say X = sin(t), then dX = cos(t) dt. The integral becomes:
∫arcsin(X) dx = ∫t cos(t) dt

Step 3: Apply the integration by parts formula
∫X^2.arcsin(X) dx = X^2 ∫arcsin(X) dx - ∫(∫arcsin(X) dx) d(X^2)

Step 4: Solve the remaining integrals
∫arcsin(X) dx = ∫t cos(t) dt

Using integration by parts:
∫t cos(t) dt = t sin(t) - ∫sin(t) dt

∫sin(t) dt = -cos(t)

Therefore:
∫t cos(t) dt = t sin(t) + cos(t)

Step 5: Final result
∫X^2.arcsin(X) dx = X^2(t sin(t) + cos(t)) - ∫(t sin(t) + cos(t)) d(X^2)

Since X = sin(t),
∫X^2.arcsin(X) dx = sin^2(t)(t sin(t) + cos(t)) - ∫(sin^2(t)(t sin(t) + cos(t))) dX

Therefore, the final answer is:
∫X^2.arcsin(X) dx = X^2(sin^2(t)(t sin(t) + cos(t))) - ∫(sin^2(t)(t sin(t) + cos(t))) dX