# indefinite integral

• Apr 30th 2010, 02:46 AM
chrizzle
indefinite integral

Cheers,
• Apr 30th 2010, 02:54 AM
Prove It
Quote:

Originally Posted by chrizzle

Cheers,

Use trigonometric substitution.

Let $x = 3\tan{\theta}$ so that $dx = 3\sec^2{\theta}\,d\theta$.

Also note that $\theta = \arctan{\frac{x}{3}}$.

So the integral becomes:

$\int{\frac{1}{x^2 + 9}\,dx} = \int{\frac{1}{(3\tan{\theta})^2 + 9}\,3\sec^2{\theta}\,d\theta}$

$= \int{\frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9}\,d\theta}$

$= \int{\frac{3\sec^2{\theta}}{9(\tan^2{\theta} + 1)}\,d\theta}$

$= \int{\frac{\sec^2{\theta}}{3\sec^2{\theta}}\,d\the ta}$

$= \int{\frac{1}{3}\,d\theta}$

$= \frac{1}{3}\theta + C$

$= \frac{1}{3}\arctan{\frac{x}{3}} + C$.
• Apr 30th 2010, 03:04 AM
chrizzle
Great, thanks

1. Does sec^2=tan^2 +1 and how do we know this?
2. Is there any way to be sure what to use instead of x in trigometric substitution?
• Apr 30th 2010, 03:08 AM
Prove It
Quote:

Originally Posted by chrizzle
Great, thanks

1. Does sec^2=tan^2 +1 and how do we know this?
2. Is there any way to be sure what to use instead of x in trigometric substitution?

1. Yes it does.

You should know the Pythagorean Identity

$\sin^2{\theta} + \cos^2{\theta} = 1$.

If you divide both sides by $\cos^2{\theta}$ you find

$\frac{\sin^2{\theta}}{\cos^2{\theta}} + \frac{\cos^2{\theta}}{\cos^2{\theta}} = \frac{1}{\cos^2{\theta}}$

$\tan^2{\theta} + 1 = \sec^2{\theta}$.

2. If you prefer to use a table of integrals, then you can use the formula

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

As you can see, this result is found using trigonometric substitution.
• Apr 30th 2010, 03:10 AM
chrizzle
actually please ignore the first question