Solving Integrals with Trig Substitution - 1/(25-x^2)

[Bryon]

Hi Everyone!
I just need some guidance on this problem. I seem to have trouble what integration technique I need to use on integrals of this type.

Homework Statement

integrate 1/(25-x^2)

Homework Equations

sqrt(a^2-u^2)
arcsin(u/a)

The Attempt at a Solution

Would I be able to use trig substitution for this type of integral? It almost fits with the arcsin but the square root is missing.

 

[snipez90]

Try partial fractions since 25-x^2 = (5-x)^2.

 

[Mark44]

Try partial fractions since 25-x^2 = (5-x)^2.

Partial fractions is the easiest way to go, but 25 - x^2 = (5 - x)(5 + x), which is not equal to (5 - x)^2.

 

[Bryon]

Ok, I believe I figured it out.

I used trig substitution

set x = 5sinx
dx/dθ = 5cosx -> dx = 5cosx dθ

I got this... int 1/(25-(5sinx)^2) 5cosx dθ which is easy to reduce.