# How To Find The Integral of Sin x dx?

1. Jul 17, 2013

### vee6

How to find the integral of sin x dx?

$$\int sin x dx=?$$

2. Jul 17, 2013

### LCKurtz

Do you know what the derivative of $\cos x$ is?

3. Jul 17, 2013

### Staff: Mentor

Do you know a function whose derivative is sin(x)? That would be the antiderivative you want.

4. Jul 17, 2013

### Mandelbroth

To find it? There are two ways I can think of:

1) Recognize that $\frac{d}{dx}[-\cos(x)+C]=\sin(x)$

2) See that $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, and then integrate the complex exponentials.

5. Jul 18, 2013

### vee6

What is this?

$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

6. Jul 18, 2013

### guest1234

Last edited by a moderator: May 6, 2017
7. Jul 18, 2013

### micromass

Staff Emeritus
How do you find $\int e^{ix}dx$ in the first place? Without using the form as a sine or cosine?

8. Jul 19, 2013

### lukka

$$\int e^{ix}~dx=i^3e^{ix} + c$$

Last edited: Jul 19, 2013
9. Jul 19, 2013

### micromass

Staff Emeritus
How do you prove this though, without knowing what the integral of sin(x) is?

10. Jul 19, 2013

### lukka

since i^3 (i) = i^4 = 1

dy/dx [i^3(e^ix)] = e^ix

11. Jul 19, 2013

### micromass

Staff Emeritus
Ok, how do you know what the derivative of $e^{ix}$ is without using sines and cosines?

12. Jul 19, 2013

### lukka

I'm not sure you can for anything but the most vigorous proof, but one will often know from experience that the exponential function that which is a weighted sum of the sine and cosine, equals it's own derivative. Maybe someone else could provide an answer to this..

Last edited: Jul 19, 2013
13. Jul 19, 2013

### micromass

Staff Emeritus
Sure, the derivative of $e^x$ is $e^x$. But that's only for real $x$. That's my entire point. To prove the relation also for complex $x$, you need to use sines and cosines to begin with. So using complex exponentials for this problem requires you to assume something that you want to prove!

14. Jul 19, 2013

### lukka

Yes i understand what you are saying. That's a valid point you have made. So how would one go about that Micro? expand the complex exponential into the trigonometric form, separate the terms and integrate?

Last edited: Jul 19, 2013
15. Jul 20, 2013

### Mandelbroth

I'd use a Taylor series. As usual, micro's point is valid. However, if we assume the exponential function is holomorphic in the whole complex plane (I'm presenting this without much rigorous proof because I'm feeling a little lazy right now), it follows that the derivative of the exponential function in the complex plane is the same as the derivative of its Taylor series, id est
$$\frac{d(e^{ix})}{dx}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}\right]=\sum_{n=0}^{\infty}\frac{i^{n+1}x^n}{n!}=ie^{ix}.$$

As an aside, using a Taylor series can also lead to noticing Euler's formula.

However, I was only attempting to provide an alternate method where we were free to assume differentiation of complex exponentials worked similarly to real exponentials. I was not intending to make it a debate point. I only meant to suggest an alternative method, which I personally use because, for some reason, I find it difficult to remember the derivatives of trig functions. :tongue:

16. Jul 20, 2013

### micromass

Staff Emeritus
Right. And then you need to know Taylor series. Additionally, you must be able to justify why you can exchange derivative and series. Do you think that somebody who just asked what the derivative of sin(x) is, know these things?

17. Jul 20, 2013

### Mandelbroth

Again, no. I'm assuming they know the integral of a function of the form $f(x)=e^{ax}$. Do you think someone who would ask such a thing would be doing a rigorous proof of the concept? I was providing my method of doing it, because I think it works.

By the way...I was responding to your question with that Taylor series. He didn't ask what the derivative of $\sin{x}$ was. He asked for the integral.

18. Jul 20, 2013

### micromass

Staff Emeritus
People in calculus typically only see this for real numbers $a$ and $x$. Not for complex variables.

I know your proof works. I'm just saying it's not helpful to the OP.

19. Jul 20, 2013

### Mandelbroth

I'm letting him or her assume, without proof, that integration works similarly for complex $a$. Again, the idea of the proof was for you, because you were asking how one finds the derivative of $e^{ix}$ without sines and cosines. I was also attempting to convey that, at this point, I don't think the OP is looking for a rigorous proof.

Now, let's stop bickering about complex variables and actually help the OP. Alright?

20. Jul 20, 2013

### LCKurtz

This was done in the first three replies, none of which were acknowledged by vee6 that he understood them.