Integrate: ∫ arcsin(x) dx?

Question

I'm a little rusty with integration by parts. I'm able to do the first half of this problem. Here's what i have so far:

u = arcsin x; dv = dx; du = 1/√(1-x²) dx; v = x, so

∫ u dv = uv - ∫ v du
∫ arcsin(x) dx = x*arcsin(x) - ∫ x/√(1-x²) dx
The second integral ∫ x / √(1-x²) dx is where I'm lost. Can someone continue with this integral?

omg i cant believe im missed that lol. Using the substitution u = 1-x² seemed so obvious that i didn't even think of it lol. thanks puggy

Puggy, you made a tiny error on your answer. u = 1-x², so du = -2x dx.
So the final answer is:
x*arcsin(x) + √(1-x²),
not minus.

Answer 1

You are definitely on the right track.

Use substitution.

∫ x/√(1 - x²) dx

Before I use substitution, I'll rearrange this to make it obvious.
Moving the next to the dx, we get

∫ ( 1/√(1 - x²) x dx )

Now, we use substitution.
Let u = 1 - x². Then
du = 2x dx, so
(1/2)du = x dx

Note that x dx is the tail end of our integral, so after the substitution, (1/2) du will be the tail end. Our integral becomes

∫ 1/√(u) (1/2) du

Factor out the (1/2),

(1/2) ∫ 1/√(u) du

Which is the same as

(1/2) ∫ u^(-1/2) du

And we use the reverse power rule to solve.

(1/2) 2u^(1/2) + C

u^(1/2) + C

Resubstituting back u,

(1 - x²)^(1/2) + C

That makes your final answer

x arcsin(x) - (1 - x²)^(1/2) + C

Answer 2

Integral Of Arcsin

Answer 3

This Site Might Help You.

RE:
Integrate: ∫ arcsin(x) dx?
I'm a little rusty with integration by parts. I'm able to do the first half of this problem. Here's what i have so far:

u = arcsin x; dv = dx; du = 1/√(1-x²) dx; v = x, so

∫ u dv = uv - ∫ v du
∫ arcsin(x) dx = x*arcsin(x) - ∫ x/√(1-x²) dx
The second integral ∫ x / √(1-x²) dx is...

Answer 4

I could type it all out, but there's a page right here that shows you everything you need to know... very useful:

http://calc101.com/parts_5.html

Answer 5

let u = 1 - x^2 then du = -2xdx

so -xdx/SQRT(1 - x^2) = (du/2)/SQRT(u)

I think you can do the rest

Answer 6

x.arcsinx