Trigonmetric Integration Question

[calcnd]

Hi, I'm new to this forum. I figure it would be appropriate to briefly introduce myself before asking my question.

My name is Nick. I've graduated high school and I am presently taking the year off before attending university next fall. Having a lot of free time has allowed me to begin learning Calculus II topics that I'll inevitably encounter in the future.

Anyways, on to my question(s):

$$\int {{-1}\over{cos^2(x)}} dx$$
$$= -\int {{sin^2(x) + cos^2(x)}\over{cos^2(x)}} dx$$
$$= -\int {{sin^2(x)}\over{cos^2(x)}} dx -\int {{cos^2(x)}\over{cos^2(x)}}dx$$
$$= -\int tan^2(x) dx - \int dx$$
$$= ?$$

Alas, I do not know how to properly integrate $$tan^2(x)$$

I am relatively confident, however, that my above reasoning is corect. This is because the answer to the following integral is supposed to be:

$$\int tan^2(x)dx = tan(x) - x + C$$ Meaning:

$$\int {{-1}\over{cos^2(x)}} dx$$
$$= -\int tan^2(x) dx - \int dx$$
$$= -tan(x) + x - C - x + D$$
$$= -tan(x) + K$$


Which should be the correct answer. Thus, my problem is in integrating the tangent x squared function.Thank-you in advance,

Nick

Edit: It appears this forum uses a different tex protocol than what I am used to on another forum I frequent. I'm attempting to figure this out, but I do appologize for the confusing mess above.

Edit 2: I believe I found the culprit. The textit code does not apply on this forum. :)

 

[courtrigrad]

$$\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C$$

 

[calcnd]

Ugh... now don't I feel rather stupid.

Trigonmetric identities will be the death of me.

 

[quasar987]

$$\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C$$

Yeah but $$\int \sec^2x dx$$ is precisely what he's trying to integrate. The question is, how can it be seen that tanx is a primitive of sec²x?

 

[quasar987]

I guess you're just assumed to have differentiated tanx before and you remembered that is gave sec²x.

 

[calcnd]

Yep,

I knew $${d\over{dx}} tanx = sec^2x$$. :)