How To Find The Integral of Sin x dx?

[vee6]

How to find the integral of sin x dx?

$$\int sin x dx=?$$

 

[LCKurtz]

How to find the integral of sin x dx?

$$\int sin x dx=?$$

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

 

[Mark44]

How to find the integral of sin x dx?

$$\int sin x dx=?$$

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

 

[Mandelbroth]

How to find the integral of sin x dx?

$$\int sin x dx=?$$

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.

 

[vee6]

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.

What is this?

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

 

[guest1234]

This is obtained by using http://en.wikipedia.org/wiki/Euler's_formula]Euler's[/PLAIN] formula and trigonometric identities \sin(-x)=-\sin(x) and \cos(-x)=\cos(x). By knowing that \frac{d}{dx}e^{kx}=ke^{kx}, you can work out the integral by yourself. But memorizing simple derivatives can save your time dramatically...

 

[micromass]

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

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

 

[lukka]

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

 

[micromass]

The integral of e^ix is i^3(e^ix) +c = -i(e^ix) +c

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

 

[lukka]

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

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

 

[micromass]

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

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

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

 

[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..

 

[micromass]

I'm not sure you can for anything but the most vigorous proof Micro, 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..

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!

 

[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?

 

[Mandelbroth]

Yes i understand what you are saying. That's a valid point you have made. So how would one go about that Micro?

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.

 

[micromass]

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
$$\frac{d(e^{ix})}{dx}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}\right]=\sum_{n=0}^{\infty}\frac{i^{n+1}x^n}{n!}=ie^{ix}.$$

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?

 

[Mandelbroth]

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, knows these things?

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.

 

[micromass]

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.

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.

 

[Mandelbroth]

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.

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?

 

[LCKurtz]

Now, let's ... actually help the OP. Alright?

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

 

[CompuChip]

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

I disagree. Without indication how we are supposed to find the antiderivative (we could just state it and that would also answer the question), what the level of the OP is (e.g. how have we defined sin and cos?) and how rigorously we need to prove it, I think it's a useful point that LCKurtz has made.