Trig Substitution

Printable View

September 24th 2008, 09:16 PM
Retromingent

Trig Substitution

The problem:
integrate 1/sqr(16-x^2) dx

My answer: (@ = theta) (trig substitution)
x=sin @ a = 16 dx = cos @ sqr(16-x^2) = cos @
= integrate cos @ / cos @ = 1

The book's answer:
arcsin(x/4)+c

Where did I go wrong?
September 24th 2008, 09:21 PM
Prove It

Quote:

Originally Posted by Retromingent

The problem:
integrate 1/sqr(16-x^2) dx

My answer: (@ = theta) (trig substitution)
x=sin @ a = 16 dx = cos @ sqr(16-x^2) = cos @
= integrate cos @ / cos @ = 1

The book's answer:
arcsin(x/4)+c

Where did I go wrong?

$\int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin{\frac{x}{a}} + C$ .

In this case, $a = 4$ .

So $\int \frac{1}{\sqrt{16 - x^2}}\, dx = \int \frac{1}{\sqrt{4^2 - x^2}}\, dx = \arcsin {\frac{x}{4}} + C.$

All times are GMT -8. The time now is 02:49 AM.

Copyright © 2005-2014 Math Help Forum. All rights reserved.

Copyright © 2005-2014 Math Help Forum. All rights reserved.