Beginning Trignometric Antiderivative

[calisoca]

Homework Statement

Okay, I think I'm finally getting the hang of these antiderivatives. However, I'm still stumbling some on trigonometric functions.

Find the antiderivative of $$f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}}$$

Homework Equations

$$f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}}$$

The Attempt at a Solution

1.) $$f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}}$$

2.) $$f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ \frac{cos^2{\theta}}{cos^2{\theta}}$$

3.) $$f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ 1$$

4.) Trigonometric Identity: $$\frac{1}{\cos^2{\theta}}} \ = \ \csc^2{\theta}$$

5.) $$f(\theta) \ = \ \csc^2{\theta}\ + \ 1$$

6.) $$F(\theta) \ = \ -\cot{\theta} \ + \ \theta$$Where have I gone wrong? I know the answer isn't correct, but I'm not sure what I have done wrong here?

 

[Dick]

1/cos(t)=sec(t). Not csc(t). And you probably don't want to write the antiderivative of F as F'. Call it G or something.

 

[calisoca]

Ah, crap! I don't know where I got $$\frac{1}{\cos^2{\theta}}} \ = \ \csc^2{\theta}$$ from. That's obviously not right. Thank you for pointing out my stupid mistake.

So...

1.) $$f(\theta) \ = \ \frac{1 + \cos^2{\theta}}{\cos^2{\theta}}$$

2.) $$f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ \frac{cos^2{\theta}}{cos^2{\theta}}$$

3.) $$f(\theta) \ = \ \frac{1}{\cos^2{\theta}}} \ + \ 1$$

4.) Trigonometric Identity: $$\frac{1}{\cos^2{\theta}}} \ = \ \sec^2{\theta}$$

5.) $$f(\theta) \ = \ \sec^2{\theta}\ + \ 1$$

6.) $$F(\theta) \ = \ \tan{\theta} \ + \ \theta$$


Now, would it be $$F(\theta) \ = \ \tan{\theta} \ + \ \theta \ + \ C$$ or just $$F(\theta) \ = \ \tan{\theta} \ + \ \theta$$ ?

 

[jgens]

You need to include the constant C. While it may seem trivial or non-important, you can actually contrive anti-derivatives, which, when simplified yield 1 = 0 (or some other non-sensical equality) if you neglect the constant.

 

[calisoca]

Great. Thanks to everyone for their help!