How to Integrate arccos x by Parts?

[Chadlee88]

Could som1 please help me integrate arccos x by parts. I've done examples using integration by parts but they were all some form of multiplication, ie
y = xe^x, y = x sin x etc. I'm really unsure where to start with this problem

 

[jpr0]

Try a substitution like y=\arccos(x), dx=\ldots ?
This should give you something you can integrate by parts.

 

[Chadlee88]

If i use a substitution as you suggested i get end up with two variables in my equation:

Integral (arccos(x) dx)

let y = arccos(x)

dy/dx = -1/(sqrt(1-x^2)

dx = -sqrt(1-x^2) dy

Substitute into original equation:

Integral (arccos(x) dx)

=> Integral (y * -sqrt(1-x^2)) dy <-----I have a y and and x in the equation :S what do i do now :S

 

[acm]

I = Int ( Arccosx) dx
I = xArccosx - x/Sqrt( 1- x^2) dx
I = xArccosx - 1/2 * 2 Sqrt(1-x^2) + C
I = xArccosx + Sqrt(1 - x^2) + C

 

[HallsofIvy]

If i use a substitution as you suggested i get end up with two variables in my equation:

Integral (arccos(x) dx)

let y = arccos(x)

dy/dx = -1/(sqrt(1-x^2)

dx = -sqrt(1-x^2) dy

Substitute into original equation:

Integral (arccos(x) dx)

=> Integral (y * -sqrt(1-x^2)) dy <-----I have a y and and x in the equation :S what do i do now :S

Yes, in any substitution, it is your responsibility to see that you replace every instance of the old variable with a new. If y= arccos x then x= cos y so dx= - sin y dy.
\int arccos x dx= -\int y sin y dy