How can I solve the integral of ArcSin(x)dx using Integration by parts?

[Parceirokid]

Hi, I'm from in Brasil and I need some help..
I have no sucess resolving Integral [ArcSin(x)]dx ..

Using Integration by parts, i don't kno what to do in expression 'c':

Integral [ArcSin(x)]dx = x.ArcSin(x) - Integral[x.(1 - x^2)^-1/2]
---------a---------- -----b----- ------------c-------------

How expression 'c' turns into only (1 - x^2)^1/2 ?

Thanks for all.

PS: excuse me for possible [a lot of] gramathical erros .. i don't speak english..

 

[jambaugh]

I= \int \sin^{-1}(x)dx = x \sin^{-1}(x) - \int x (1-x^2)^{-1/2}dx
Now use trig substitution:
x = \sin(t), dx = cos(t)dt
I = x\sin^{-1}(x) - \int \sin(t)\cos(t)(1-\sin^2(t))^{-1/2}dt=\int \sin(t) \frac{\cos(t)}{\cos(t)}dt
= x\sin^{-1}(x) - \int \sin(t)dt = x\sin^{-1}(x)-\cos(t)+C***
= x\sin^{-1}(x) - \cos(\sin^{-1}(x)) + C
= x\sin^{-1}(x) - \sqrt{1 - x^2} + C

Or you can use a substitution immediately:
\int \sin^{-1}(x)dx = \int t \cos(t)dt
where
x = sin(t)
and then integrate by parts:
= \int t cos(t)dt= t\sin(t) - \int \sin(t)dt
= \sin^{-1}(x) x - \int \sin(t)dt=x\sin^{-1}(x)-\cos(t)+C
which is the same as (***) in the previous try.

 

[yip]

You could also use the much easier substitution t=x^2 to evaluate \int x (1-x^2)^{-1/2}dx

 

[VietDao29]

You could also use the much easier substitution t=x^2 to evaluate \int x (1-x^2)^{-1/2}dx

And even better, you can try the substitution: t = 1 - x². :)

 

[Parceirokid]

All Right!

One more time, thanks for all.

VietDao29, I take your sugestion [so many simple and usefull] and
it works fine. Thank you.

See you..

 

[SELFMADE]

I= \int \sin^{-1}(x)dx = x \sin^{-1}(x) - \int x (1-x^2)^{-1/2}dx

Can someone explain this part please?

 

[arildno]

Use integration by parts, noting that:
\sin^{-1}(x)=1*\sin^{-1}(x)
and:
\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^{2}}}

 

[SELFMADE]

thanks