Can some one please tell me what is:

∫sinⁿx

I can't be bothered working it out with integration by parts!

P.s. can someone tell me how to get math html symbols up too!!!

sinx=(exp(ix)-exp(-ix))/2i. Use the binomial theorem to expand sin^n(x) as a sum of terms of the form cexp(ikx) (c depends on k) k will take on values from n to -n in steps of 2. You can then integrate term by term. Continue this type of manipulation in reverse (exp back to sin or cos) and get a finite trig series.

John Connellan: Try using <> instead of [] for subscripts & superscripts.

The special character codes appear as follows

e.g.

For &int;, the code is & int ; with no spaces.

You can find the integral

&int; sinⁿ(x) dx

without complex numbers by using reduction formulae.

Is this right:

&int;sin²x = 1/2(x)-1/4 (sin 2x) + C

and

&int;sin³x = -1/3(sin²x)(cos x) -2/3(cos x) + C

&int

Above Used & int without the space. It did not show as an integral sign.

Dinosaur, you didn't put the semicolon on the end.

&int

Above Used & int without the space. It did not show as an integral sign.

AD1 told me to do it without spaces! Anyway here we go again.

&int;sin(x)

I hope that worked!

What I was actually asking though is if the result of the integration above was right?!

Im gonna make this my integration 'questioning' thread if that oks.

Can some one tell me what the answer to this is?

&int;sin((x-1)*c)

Distribute the c within the brackets:

&int; sin((x - 1)c) dx = &int; sin(cx - c) dx

The general form is

&int; sin(ax + b) dx = -(1/a)cos(ax + b) + C

The phase angle is irrelevant, only the coefficient of x is relevant. So in this case, the answer is

-(1/c)cos(cx - c) + C

Thanks a lot AD1. I should really start reading up on some calculus! I only know the basics.

To do:

Integral [sin ((x-1)c)] dx,

first substitute

u = c(x-1)

Then

du/dx = c

or, in other words dx = (1/c) du. Therefore

Integral [sin(c(x-1))] dx = Integral [sin u (1/c)] du = -(1/c) cos u = -(1/c) cos (c(x-1)) + constant

How would u go about integrating this (2-wave) function?

&int; A₁Sin(wx+b)+A₂Sin(hx+d)+c dx

Use the method given above and integrate each term seperately.

Your integral is then:

-(A₁/w) cos(wx + b) - (A₂/h) cos(hx + d) + x + C.

OK sorry. Knew how to do each part seperately but forgot that in integration u CAN just do them seperately when addition is involved :rolleyes:

John Connellan

I can't be bothered working it out with integration by parts!

I'll give you a hint if you want to be a successful physicist or mathematician.

BE BOTHERED.

You sound as if you treat these problems as a chore, you're in a rush to solve them. But I bet you can sit for hours on the internet and read a whole lot of infodribble. Trust me, there is no greater satisfaction in arriving as some formula or method that someone else did 50, 100 etc years ago.
Also by doing these problems you will learn little tricks which will help you with harder problems in the future.

If you are going to be an engineer or a chemist, ignore that and ask someone for the answer, it makes us feel important

But I bet you can sit for hours on the internet and read a whole lot of infodribble.
infodribble. hahaha. that is the best description i have ever heard of the kind of stuff we get on sciforums. i like that term.

Trust me, there is no greater satisfaction in arriving as some formula or method that someone else did 50, 100 etc years ago.
except perhaps arriving at some formula that no one has ever arrived at before

Yes, but I usually make so many assumptions on the way that it is totally irrelevant to any problem

You sound as if you treat these problems as a chore, you're in a rush to solve them.

Thats true actually. Only because I read a book on calculus before and was quickly bored once I got beyond the basics. Im beginning to realise what ur talking about now though so I want to start learning again.