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

(Please see attatchment)

Cheers,

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- Apr 30th 2010, 01:46 AMchrizzleindefinite integral
Can anyone please help to find an antiderivative of this function and the method used?

(Please see attatchment)

Cheers, - Apr 30th 2010, 01:54 AMProve It
Use trigonometric substitution.

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

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

So the integral becomes:

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

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

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

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

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

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

$\displaystyle = \frac{1}{3}\arctan{\frac{x}{3}} + C$. - Apr 30th 2010, 02:04 AMchrizzle
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? - Apr 30th 2010, 02:08 AMProve It
1. Yes it does.

You should know the Pythagorean Identity

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

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

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

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

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

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

As you can see, this result is found using trigonometric substitution. - Apr 30th 2010, 02:10 AMchrizzle
actually please ignore the first question