Expert: Paul Klarreich - 5/19/2008

Question
I need the proofs of
intergration of du/squareroot(a^2-u^2)
intergration of du/a^2+u^2
intergration of du/u*squareroot(u^2-a^2)
Thank you~

Answer
Questioner:   Heidi
Category:  Advanced Math
Private:  No
 
Subject:  proofs for integrals involving inverse trig functions
Question:  I need the proofs of
integration of du/squareroot(a^2-u^2)
integration of du/a^2+u^2
integration of du/u*squareroot(u^2-a^2)
Thank you~
---------------------------
Hi, Heidi,

1. It's called INTE GRATION. (only one 'R')

2. Every integral formula is the reverse of a differentiation formula.

One way to do it:  To prove that

{     du
| --------------- = arcsin(u/a), as you see it in the book,
} sqrt(a^2 - u^2)

Let y = arcsin(u/a) and prove that
          du
dy = ---------------
    sqrt(a^2 - u^2)      

Prove it by differentiation:

y = arcsin(u/a)

u/a = sin y.

u = a sin y.

Differentiate implicitly:

du = a cos y dy, and

dy = du/(a cos y)

Now do some trig:

a^2 cos^2(y) = a^2(1 - sin^2(y))

But sin y = u/a:

a^2 cos^2(y) = a^2(1 - u^2/a^2)

a^2 cos^2(y) = a^2 - u^2

Some algebra:

a cos y = sqrt(a^2 - u^2)

So you have:

dy = du/sqrt(a^2 - u^2),

which is the formula.

..................................

3. Don't like that approach?  Try integration by TRIG SUBSTITUTION, which you may or may not have learned:

To integrate

{     du
| ---------------
} sqrt(a^2 - u^2)

Use a right triangle in which:

The acute angle is t (theta?)

a is the hypotenuse
u is the leg opposite angle t.
sqrt(a^2 - u^2) is the leg adjacent to angle t.

Then  
sin t = u/a, and  u = a sin t [the subst]
du = a cos t dt
sqrt(a^2 - u^2) = a cos t

Now you get:
{     du
| --------------- =
} sqrt(a^2 - u^2)

{    a cos t dt
| -------------- =
}  a cos t

{   
| dt = t
}

and since u = a sin t,

sin t = u/a, and  t = arcsin(u/a),

which is your answer.

.............................................

I think, after all this, you should bve able to handle the others with no trouble.  If you can't, let me know.