	AllExperts > Experts
	Search

About Paul Klarreich
Expertise
I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience
I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

You are here: Experts > Science > Mathematics > Advanced Math > Integral of inverse trig functions

Topic: Advanced Math

Expert: Paul Klarreich
Date: 5/19/2008
Subject: Integral of inverse trig functions

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.

Add to this Answer Ask a Question

About Us | Advertise on This Site | User Agreement | Privacy Policy | Help
Copyright © 2008 About, Inc. About and About.com are registered trademarks of About, Inc. The About logo is a trademark of About, Inc. All rights reserved.