• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Indefinite Integral (1 Viewer)

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Using complex numbers or induction (I prefer the complex number method), you can prove that:



So the question at hand is the case for n=9:




Integrate both sides w.r.t x gives:

 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Oh I forgot to include the dx and the +C, but you get the idea.
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,253
Gender
Male
HSC
N/A
That's cool. Never would've thought of it.

Still unable to rep you - seems like I haven't been liberal enough with my repping.
 
Last edited:

juantheron

Active Member
Joined
Feb 9, 2012
Messages
259
Gender
Male
HSC
N/A
Thanks Mod. for Nice solution.

But How can I prove the Given Result using Complex no.



Thanks
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Thanks Mod. for Nice solution.

But How can I prove the Given Result using Complex no.



Thanks
Ugh my computer froze as I was typing out the solution on LaTeX. I'll re-do it tomorrow if I have time.

But I'll get you started. Not sure how nicely De Moivre's Theorem turns out if you use the cos(x) + isin(x) = cis(x) to prove the question, but it can be done without too much difficulty if you use the complex exponential definitions of the sine function.



Try it yourself! If you are at a level to be attempting these types of integrals, I am sure you are capable of doing this.

But then again, this definition of the sine function is not within the HSC Syllabus (but good to know anyway!).

A more elementary (but consequently tougher on the algebra) method of proving this would be to realise that the series sin(x) + sin(3x) + sin(5x) + ... + sin(2k-1)x is a Trigonometric Sum with the angles in Arithmetic Progression such that a = x, and d = 2x.

There is also a closed form for such a thing. If I recall correctly, the formula is as follows:

 

juantheron

Active Member
Joined
Feb 9, 2012
Messages
259
Gender
Male
HSC
N/A
Thanks Drongoski and Moderator

actually it is from my assignment

Thanks
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Fabulous proof Drongoski! Thank you very much.

Must spread rep.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top