How do I use the complex exponential form of sin(x) for integration?

[Al-Layth]

I am told this identity:
[math]\sin(x) = \frac{e^{ix} - e^{-ix} }{2i}[/math]
How can I use this identity to integrate sin(x) ?

[math]\int{\sin(x)}dx \longrightarrow \int{\frac{e^{ix} - e^{-ix} }{2i}}dx \longrightarrow \frac{1}{2i}\int{e^{ix}}dx - \frac{1}{2i} \int{e^{-ix}}dx[/math]
I don't know how integration works if there are imaginary numbers involved.

plz help thank

 

[Zermelo]

I am told this identity:
[math]\sin(x) = \frac{e^{ix} - e^{-ix} }{2i}[/math]
How can I use this identity to integrate sin(x) ?

[math]\int{\sin(x)}dx \longrightarrow \int{\frac{e^{ix} - e^{-ix} }{2i}}dx \longrightarrow \frac{1}{2i}\int{e^{ix}}dx - \frac{1}{2i} \int{e^{-ix}}dx[/math]
I don't know how integration works if there are imaginary numbers involved.

plz help thank

Imaginary numbers are just constants ? The rules for integration are the same as with real constants, for example, what would you do if it was 4x instead of ix?

 

[Al-Layth]

thanks

that worked

one more question: Although this time I could recognise the expression for -cos(x) . Is there a general way of converting a complex algebraic expression into a form with no imaginary numbers? (given that you know the expression is real despite containing imaginary terms, just as I knew the integral of sin(x) should be real)

 

[Dr.Peterson]

thanks

that worked

one more question: Although this time I could recognise the expression for -cos(x) . Is there a general way of converting a complex algebraic expression into a form with no imaginary numbers? (given that you know the expression is real despite containing imaginary terms, just as I knew the integral of sin(x) should be real)

You know that [imath]e^{ix}=\cos(x)+i\sin(x)[/imath], right?

Put that into the expression you got, and simplify. If you're right about the answer being real, the i's should go away.

 

[Al-Layth]

Ohhh.
thank u so much

You know that [imath]e^{ix}=\cos(x)+i\sin(x)[/imath], right?

Put that into the expression you got, and simplify. If you're right about the answer being real, the i's should go away.

 

[Steven G]

Please use equal signs instead of arrows.

 

[Julicjn]

Please tell me how to answer an existing topic?
Maybe I'm not writing correctly?
Please help.
Thank you.

 

[Al-Layth]

Please use equal signs instead of arrows.

noted

 

[Steven G]

one more question: Although this time I could recognise the expression for -cos(x) . Is there a general way of converting a complex algebraic expression into a form with no imaginary numbers? (given that you know the expression is real despite containing imaginary terms, just as I knew the integral of sin(x) should be real)

Maybe I am not understanding your question, but if you're given \(\displaystyle \int \dfrac {e^{ix} - e^{-ix}}{2i}dx\), then notice that \(\displaystyle \dfrac {e^{ix} - e^{-ix}}{2i} = sin x\) and \(\displaystyle \int \dfrac {e^{ix} - e^{-ix}}{2i}dx = \int sinx dx = -\cos x + c \)