Find the general indefinite integral

In summary, the conversation is about finding the general indefinite integral of \int\frac{\sin{x}}{1-sin^2{x}}dx and one person is asking for help with trigonometry. Another person suggests using the substitution method and provides a step-by-step explanation of how to solve the integral. The original person thanks them for their help and plans to revisit the solution after getting some rest.
  • #1
calisoca
28
0

Homework Statement



I wasn't really sure where to post this, as I don't need help understanding the integration. I need help with the trigonometry!

That being said, here is the problem.

Find the general indefinite integral of [tex] \int\frac{\sin{x}}{1-sin^2{x}}dx [/tex]



Homework Equations



[tex] \int\frac{\sin{x}}{1-sin^2{x}}dx [/tex]



The Attempt at a Solution



I've tried starting from [tex] \cos^2{x} + \sin^2{x} = 1 [/tex].

That gets me to [tex] \frac{\sin{x}}{\cos^2{x}} [/tex], but I'm not sure what good that really does me?

I'm a little stuck here, and some help would be much appreciated!
 
Physics news on Phys.org
  • #2


Well, what's the derivative of cos(x)?
 
  • #3


gabbagabbahey said:
Well, what's the derivative of cos(x)?

[tex] \sin{x} + C [/tex]

I apologize, but I'm still not following.
 
  • #4


calisoca said:
[tex] \sin{x} + C [/tex]

I apologize, but I'm still not following.

Ermm isn't it just -sin(x)...I asked for the derivative, not the integral...Anyways, that tells you that:

[tex]\frac{d}{dx}\cos x=-\sin x \implies \sin x dx =-d(\cos x)[/tex]

[tex]\implies \int\frac{\sin{x}}{cos^2{x}}dx = \int\frac{-d(\cos x)}{cos^2{x}}[/tex]

Or, writing [tex]u\equiv \cos x[/tex]

[tex]\int\frac{-du}{u^2}[/tex]
 
  • #5


I apologize. It's been a very long day, and I'm crazy tired. I meant to give you the derivative.

The problem is that we haven't done anything like what you just described, so I'm having a hard time figuring it out. It looks like you used some sort of substitution. I've looked ahead, and that comes in the next chapter.

Maybe I don't need to know anything about the next chapter to understand what you just did. Honestly, I'm probably not seeing it, because I'm technically medically brain dead right now.

I'll brew on this and see if I don't come to tomorrow.

At any rate, I greatly appreciate your patience and help!
 
  • #6


Look at it this way, this is not very rigorous but here goes:

You have 3 terms: [tex] sin(x) \cdot \frac{1}{cos^2(x)} \cdot dx [/tex]. So now let's do something called u-subst.

If we let [tex] u = cos(x) [/tex] then [tex] du/dx = -sin(x) \Rightarrow -du = sin(x) dx[/tex].

So now let's look at our equation, we can group [tex] sin(x) \cdot \frac{1}{cos^2(x)} \cdot dx = \frac{1}{cos^2(x)} \cdot sin(x) \cdot dx [/tex].

Well comparing that to our "clever" subst. we see that's the same as [tex] \frac{1}{u^2} \cdot -du [/tex].

So using the subst. [tex] u = cos(x) [/tex] we get that

[tex] \int \frac{sin(x)}{cos^2(x)} dx = -\int \frac{1}{u^2} du [/tex]

as Gabba said.
 
  • #7


Thank you very much.

I'm at that point where it's obvious this makes sense. If I hadn't been up for the last 48 hours, then I wouldn't even need to ask the question! Ha.

Nonetheless, yes, I see superficially that this will make sense later, but my brain will not let me comprehend it right now. I honestly didn't expect a response back on this question so quickly. I thought I'd just post it and check tomorrow after I had some sleep.

Thank you very much, both of you, for your patience and your replies. I am going to let this sit on my brain, and then revisit it tomorrow. If I have further questions, I'll let you know, but with those two great explanations, there's no reason why I should have any more questions, really.

Thank you again.
 

Related to Find the general indefinite integral

1. What is the general indefinite integral?

The general indefinite integral, also known as the antiderivative, is the reverse process of differentiation in calculus. It is a function that, when differentiated, gives back the original function.

2. How do I find the general indefinite integral?

To find the general indefinite integral, you must use integration rules and techniques such as u-substitution, integration by parts, or trigonometric substitution. It also requires knowledge of basic algebra and calculus principles.

3. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral does not have limits and gives a general function as a solution.

4. Can every function be integrated?

No, not all functions have a closed-form antiderivative. Some functions, such as the exponential function, have no general indefinite integral and require special techniques or approximations to be integrated.

5. How do I know if my answer for the general indefinite integral is correct?

You can check your answer by taking the derivative of the obtained function. If the derivative is equal to the original function, then your answer is correct. You can also use online integration calculators to verify your solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
454
  • Calculus and Beyond Homework Help
Replies
15
Views
871
  • Calculus and Beyond Homework Help
Replies
11
Views
765
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
921
  • Calculus and Beyond Homework Help
Replies
27
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
574
Replies
2
Views
360
  • Calculus and Beyond Homework Help
Replies
6
Views
677
Back
Top