Homework Help: A SIMPLE integral, but I want a nice way to solve.

1. Aug 24, 2011

1MileCrash

1. The problem statement, all variables and given/known data

p(t) = cos(t) + (1)/(cos^2(t)

Find antiderivative.

2. Relevant equations

3. The attempt at a solution

Alright, I'm just starting calc 2. I thoroughly covered a lot of integration involving u substitution, and we have not started that in calc 2 yet. U substitution is so ingrained into my mind right now.

Wolfram gives sin(t) + tan(t) which I agree with by quickly going through differentiation, but what rules of integration would you apply? I'm guessing some algebra or trig identity?

2. Aug 24, 2011

Yuqing

That p(t) is exactly equal to cos(x) + sec2(x). This can be done simply with direct integration.

3. Aug 25, 2011

mstud

You can also solve it like this:

$$p(t) = cos(t) + \frac 1{cos^2(t)}$$

Substitute u=cos(t):

$$p(t)=u+u^{-2}$$$$\int p(t)= \int u+u^{-2}=...$$

That's a possible way if you wan't to use substitution.

4. Aug 25, 2011

SammyS

Staff Emeritus
If u=cos(t) , then du = -sin(t) dt . Furthermore, $\displaystyle \sin(t) = \sqrt{1-\cos^2(t)} = \sqrt{1-u^2}\,.$

Therefore, $\displaystyle \int p(t)\,dt= -\int \left(u+u^{-2}\right)\sqrt{1-u^2}\,du\,.$

So, that's probably not the way to go.

See Yuqing's post for a straight forward method.

Last edited: Aug 25, 2011
5. Aug 25, 2011

HallsofIvy

Sorry but that's invalid specifically because that is a "substitution" but you ignored the "dx" and "du". If $u= cos(t)$ then $du= -sin(t)dt$. To integrate $\int (cos(t)+ 1/cos(t)) dt$ and letting u= cos(t) you would have $\int (u+ 1/u^2)dx$ and substituting for dx will give you a very complex integral.

6. Aug 25, 2011

gb7nash

Here's a solution. However, it does require the circular reasoning that the derivative of tan is sec2. If you can ignore that though, it's legit:

$$\int sec^{2}x dx = \int \sec^{2}x \frac{\tan x}{\tan x} dx$$

Letting u = tanx, du = sec2x dx, we get:

$\int \frac{u}{u} du = \int du =$ ...

7. Aug 26, 2011

HallsofIvy

But the "tan x/tan x" is irrelevant. All you are doing is saying $d(tan(x)= sec^2(x)dx$ so $\int sec^2(x)dx= \int d(tan x)= tan(x)+ C$.

8. Aug 26, 2011

SammyS

Staff Emeritus
I had a student who would routinely carry out his substitutions this far. At first it caught me off guard. Then I caught on to what he was doing. He would pick u so that du/dx was equal to the whole integrand, so that he would get: $\displaystyle \int\,\frac{du}{dx}\,dx=\int\,du=u+C\,.$

From that time on, I would often refer to this as "Huang's method" , crediting that student.

Last edited: Aug 26, 2011
9. Aug 26, 2011

gb7nash

Good point. I didn't realize this.