# Constructing Antiderivatives!!

• Apr 23rd 2008, 11:53 AM
mathlete
Constructing Antiderivatives!!
Find the general antiderivative of the function below. Use theta for θ.

http://www.webassign.net/www31/symIm...49fc1a6e16.gif

P(θ) = + C

need help with another one please, just dont know where to begin, any help would be fantastic, thanks
• Apr 23rd 2008, 11:58 AM
o_O
Some hints:
$\frac{d}{dx} \sec \theta = \sec \theta \tan \theta$
$\frac{d}{dx} \tan \theta = \sec^{2} \theta$

Can you see how these apply to your integral? Basically you're going backwards from derivatives you learned in differentiation.
• Apr 24th 2008, 04:50 PM
bluejewballs
Quote:

Originally Posted by o_O
Some hints:
$\frac{d}{dx} \sec \theta = \sec \theta \tan \theta$
$\frac{d}{dx} \cot \theta = -\csc^{2} \theta$

Can you see how these apply to your integral? Basically you're going backwards from derivatives you learned in differentiation.

Hey i have a problem similar to this and I dont see how that applies to the question.
• Apr 24th 2008, 04:51 PM
Mathstud28
Quote:

Originally Posted by bluejewballs
Hey i have a problem similar to this and I dont see how that applies to the question.

the first part would be because the derivative of the integral or vice versa is the original function
• Apr 24th 2008, 04:53 PM
o_O
Let me colour code this:

$p(\theta) = {\color{red} \sec \theta \tan \theta } + {\color{blue}\frac{6}{\cos^{2} \theta}}$

$\frac{d}{dx} \sec \theta = {\color{red} \sec \theta \tan \theta }$
$\frac{d}{dx} \tan \theta = {\color{blue}sec^{2} \theta}$**

** Should've been this one in the first post