Find the general indefinite integral

[calisoca]

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 $$\int\frac{\sin{x}}{1-sin^2{x}}dx$$

Homework Equations

$$\int\frac{\sin{x}}{1-sin^2{x}}dx$$

The Attempt at a Solution

I've tried starting from $$\cos^2{x} + \sin^2{x} = 1$$.

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

I'm a little stuck here, and some help would be much appreciated!

 

[gabbagabbahey]

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

 

[calisoca]

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

$$\sin{x} + C$$

I apologize, but I'm still not following.

 

[gabbagabbahey]

$$\sin{x} + C$$

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:

$$\frac{d}{dx}\cos x=-\sin x \implies \sin x dx =-d(\cos x)$$

$$\implies \int\frac{\sin{x}}{cos^2{x}}dx = \int\frac{-d(\cos x)}{cos^2{x}}$$

Or, writing $$u\equiv \cos x$$

$$\int\frac{-du}{u^2}$$

 

[calisoca]

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!

 

[NoMoreExams]

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

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

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

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

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

So using the subst. $$u = cos(x)$$ we get that

$$\int \frac{sin(x)}{cos^2(x)} dx = -\int \frac{1}{u^2} du$$

as Gabba said.

 

[calisoca]

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.