Can anyone please help to find an antiderivative of this function and the method used?
(Please see attatchment)
Cheers,
Printable View
Can anyone please help to find an antiderivative of this function and the method used?
(Please see attatchment)
Cheers,
Use trigonometric substitution.Quote:
Originally Posted by chrizzle
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:
$\displaystyle \int{\frac{1}{x^2 + 9}\,dx} = \int{\frac{1}{(3\tan{\theta})^2 + 9}\,3\sec^2{\theta}\,d\theta}$
$\displaystyle = \int{\frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9}\,d\theta}$
$\displaystyle = \int{\frac{3\sec^2{\theta}}{9(\tan^2{\theta} + 1)}\,d\theta}$
$\displaystyle = \int{\frac{\sec^2{\theta}}{3\sec^2{\theta}}\,d\the ta}$
$\displaystyle = \int{\frac{1}{3}\,d\theta}$
$\displaystyle = \frac{1}{3}\theta + C$
$\displaystyle = \frac{1}{3}\arctan{\frac{x}{3}} + C$.
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.Quote:
Originally Posted by chrizzle
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.
actually please ignore the first question
