Integrating arcsin.. using parts??

[Melissaherman]

Hi! I've got an integration question that is kicking my butt right now...
It is to integrate inverse sin of x (in other words, sin^-1 x dx)
My prof told me to use integration by parts... so I tried that, and after using parts once, had:
∫ sin^-1x dx = xsin^-1x - ∫ (1 / {sqrt(1-x^2)}) x


So I tried parts again, and ended up with:

∫ sin^-1x dx = xsin^-1x – xsin^-1x - ∫ sin^-1x

Sooo then I thought, hey, this is a wrap-around integral! But, since I have xsin^-1x - xsin^-1x, it would end up being:

∫ sin^-1x dx = ∫ sin^-1x dx

Which makes absolutely no sense. I checked my signs in my work, and I'm pretty sure that is right... what am I doing wrong??

Thanks for any help!

 

[pka]

You should know about this link:
http://integrals.wolfram.com/index.jsp

 

[Unco]

Hi! I've got an integration question that is kicking my butt right now...
It is to integrate inverse sin of x (in other words, sin^-1 x dx)
My prof told me to use integration by parts... so I tried that, and after using parts once, had:
∫ sin^-1x dx = xsin^-1x - ∫ (1 / {sqrt(1-x^2)})x dx <- This last integral can be evaluated using a standard substitution.

 

[Count Iblis]

Hi! I've got an integration question that is kicking my butt right now...
It is to integrate inverse sin of x (in other words, sin^-1 x dx)
My prof told me to use integration by parts... so I tried that, and after using parts once, had:
∫ sin^-1x dx = xsin^-1x - ∫ (1 / {sqrt(1-x^2)}) x


So I tried parts again, and ended up with:

∫ sin^-1x dx = xsin^-1x – xsin^-1x - ∫ sin^-1x

Sooo then I thought, hey, this is a wrap-around integral! But, since I have xsin^-1x - xsin^-1x, it would end up being:

∫ sin^-1x dx = ∫ sin^-1x dx

Which makes absolutely no sense. I checked my signs in my work, and I'm pretty sure that is right... what am I doing wrong??

Thanks for any help!

Well, what's the derivative of sqrt[1-x^2] ?

 

[soroban]

Hello, Melissa!


We have: \(\displaystyle \L\:\int\)\(\displaystyle \sin^{-1}(x)\,dx\)

By parts: \(\displaystyle \:\begin{array}{ccc}u\,=\,\sin^{-1}(x) & \;\qquad\; & dv\,=\,dx \\
du\,=\,\frac{dx}{\sqrt{1\,-\,x^2}} & \;\qquad\: & v\,=\,x\end{array}\)

Then we have: \(\displaystyle \L\:x\cdot\sin^{-1}(x) \,-\,\int\frac{x\,dx}{\sqrt{1\,-\,x^2}}\)

. . This integral can be handled with: \(\displaystyle \,u\:=\:1\,-\,x^2\)

 

[Melissaherman]

Hello, Melissa!


We have: \(\displaystyle \L\:\int\)\(\displaystyle \sin^{-1}(x)\,dx\)

By parts: \(\displaystyle \:\begin{array}{ccc}u\,=\,\sin^{-1}(x) & \;\qquad\; & dv\,=\,dx \\
du\,=\,\frac{dx}{\sqrt{1\,-\,x^2}} & \;\qquad\: & v\,=\,x\end{array}\)

Then we have: \(\displaystyle \L\:x\cdot\sin^{-1}(x) \,-\,\int\frac{x\,dx}{\sqrt{1\,-\,x^2}}\)

. . This integral can be handled with: \(\displaystyle \,u\:=\:1\,-\,x^2\)

ooh I see... I actually tried that at first, but to be honest I wasn't really sure what the integral of 1/sqrt(u) would be.... But I check out that website that does integrals for you, and now it makes sense.
Thank you so much for your help!