# Math Help - indefinite integral

1. ## indefinite integral

Can anyone please help to find an antiderivative of this function and the method used?

Cheers,

2. Originally Posted by chrizzle
Can anyone please help to find an antiderivative of this function and the method used?

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:

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

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

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

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

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

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

$= \frac{1}{3}\arctan{\frac{x}{3}} + C$.

3. Great, thanks

Just 2 follow up questions:
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?

4. Originally Posted by chrizzle
Great, thanks

Just 2 follow up questions:
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.

5. actually please ignore the first question