How do you find the integral of arcsin(x)dx?

Question

I know you have to do integration by parts, but I'm stuck at the end... Note: INT = integral

So far I have:

INT arcsin(x) =INT sin^(-1)x

u = sin^(-1)x
du = 1/(sqrt(1+x^2))

dv = dx
v = x

so then by the formula for integration by parts uv - INTvdu
which gives me xsin^(-1)x - INT(x/sqrt(1+x^2)) so then you have to do substitution to find the integral of that... and I did

u = 1 + x^2
du = 2xdx

So I plug it all in and do all that good stuff and I end up with

xsin^(-1)x - sqrt(sin^(-1)x)

But it's not right... Please help me with what I am doing wrong!!

Answer 1

Looks like you started it right. I looked up the derivative of sinˉ¹(x), though, and it should be 1/√(1 - x²), not 1/√(1 + x²).

∫ sinˉ¹(x) dx

u = sinˉ¹(x)
du = 1/√(1 - x²) dx

dv = dx
v = x

∫ sinˉ¹(x) dx = x sinˉ¹(x) - ∫ x/√(1 - x²) dx

w = 1 - x²
dw = -2x dx
-1/2 dw = x dx

You were right down to here...but I don't see how you ended up with sinˉ¹(x) again from the right-hand integral....

∫ sinˉ¹(x) dx = x sinˉ¹(x) + 1/2∫ 1/√w dw =

∫ sinˉ¹(x) dx = x sinˉ¹(x) + 1/2∫ w^(-1/2) du =

The integral of w^(-1/2) is 2w^(1/2), or 2√w:

∫ sinˉ¹(x) dx = x sinˉ¹(x) + 1/2 [2√w] + C =

Then re-substitute...maybe because you re-used "u", you substituted in the wrong u?

∫ sinˉ¹(x) dx = x sinˉ¹(x) + √(1 - x²) + C

Answer 2

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