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

[1MileCrash]

Homework Statement

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

Find antiderivative.

Homework Equations

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?

 

[Yuqing]

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

 

[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.

 

[SammyS]

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 want to use substitution.

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.

 

[HallsofIvy]

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.

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.

 

[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:

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

 

[HallsofIvy]

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:

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

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

 

[SammyS]

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:
...

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

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.

 

[gb7nash]

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

Good point. I didn't realize this.