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