Integrating trig equations

[D'Nalor]

Quick question, how do I integrate this?

 

~sin(2x)cos(3x)sin(4x)dx

 

I can't figure out how to do it. I've managed to get an answer using

cos(nx)=(z^n+z^-n)/2 and sin(nx)=(z^n-z^-n)/2i, but somehow I get the feeling that isn't how I'm supposed to do it...

I am familiar with inegration by parts, ~F(x)=~G(u)u' [and another one like the previous that I can't remember how to write], double angle rules, and naturally 1=sin(x)^2+cos(x)^2

I still can't figure out how to do it though, given that we're supposed to be able to do it using only those.

I think I remember reading somewhere a rule that I think was sin(a+b)=sin(a)-sin(b), but I don't think it will really help in this one.

Please help me.

 

note: ~ represents integration.

[ydoaPs]

It's been a while since I've done much math, but it might help if you use our built-in LaTex system.

An example:

[math]\int{f(x)}dx=\lim_{\Delta{x}\rightarrow{0}}\sum{f(x)}\Delta{x}[/math]

 

If you click on it, you'll get a popup that displays the code. You can also get the code by using the quote function.

[D'Nalor]

ok then. Formula is to integrate is

 

[math]\int{sin(2x)cos(3x)sin(4x)dx}[/math]

[Cap'n Refsmmat]

There's a rule that [imath]\sin \alpha \sin \beta = \frac{1}{2} (\cos(\alpha-\beta)-\cos(\alpha+\beta))[/imath]. That may help.

 

Also, [imath]\cos \alpha \cos \beta = \frac{1}{2} (\cos(\alpha-\beta)+\cos(\alpha+\beta))[/imath], I think.

 

I hate trig identities.

[ydoaPs]

There's a rule that [imath]\sin \alpha \sin \beta = \frac{1}{2} (\cos(\alpha-\beta)-\cos(\alpha+\beta))[/imath]. That may help.

 

I forgot that one. Sorry, it's been a few years since trig. This is definately the one needed. The use of brackets may also help.

 

[math]\int\sin(2x)[\sin(4x)\cos(3x)]dx[/math] in conjunction with the above quoted trig identity leads us to:

 

[math]\int\sin(2x){(\frac{1}{2}(\cos(x)-\cos(7x))}dx[/math]

which leads to

[math]\int(\frac{1}{2}\sin(2x)\cos(x)-\frac{1}{2}\sin(2x)\cos(7x))dx[/math]

using the same trig identity gives

[math]\int((\frac{1}{4}\cos(0)+\frac{1}{4}\cos(3x))-(\frac{1}{4}\cos(0)+\frac{1}{4}\cos(10x))dx[/math]

which reduces to

[math]\frac{1}{4}(\int\cos(3x)dx-\int\cos(10x)dx)[/math]

 

Now, that is much easier.

 

 

Like I said, it's been a while, but I got:

[hide](1/12)sin(3x)-(1/40)sin(10x)[/hide]

Edited June 14, 2010 by ydoaPs

[D'Nalor]

I forgot that one. Sorry, it's been a few years since trig. This is definately the one needed. The use of brackets may also help.

 

[math]\int\sin(2x)[\sin(4x)\cos(3x)]dx[/math] in conjunction with the above quoted trig identity leads us to:

 

[math]\int\sin(2x){(\frac{1}{2}(\cos(x)-\cos(7x))}dx[/math]

 

Urm... I'm affraid you lost me there. Is there some other rule in there I don't know about? the quoted rule doesn't use sin and cos at the same time. Did you mean to bracket it like this?

 

[math]\int[\sin(2x)\sin(4x)]\cos(3x)dx[/math]

 

That would allow you to make that substitution. Certainly from what you put down, the answer is right. Using the corectly bracketed version, that would make it...

 

[math]\int[cos(-2x)-cos(6x)]\cos(3x)dx[/math]

 

[math]\int cos(-2x)cos(3x)dx -\int cos(6x)cos(3x)dx[/math]

 

...which doesn't make it too much easier...

 

Are you sure that rule is right? from

 

[imath]

\sin \alpha \sin \beta = \frac{1}{2} (\cos(\alpha-\beta)-\cos(\alpha+\beta))

[/imath]

 

[imath]

\sin \beta \sin \alpha = \frac{1}{2} (\cos(\beta-\alpha)-\cos(\beta+\alpha))

[/imath]

 

and as

 

[math]\sin \alpha \sin \beta = \sin \beta \sin \alpha[/math]

 

but I don't think this is right...

 

[math]\frac{1}{2} (\cos(\beta-\alpha)-\cos(\beta+\alpha)) = \frac{1}{2} (\cos(\alpha-\beta)-\cos(\alpha+\beta))[/math]

 

Just so I can show an example of

 

[math]sin(\frac{\pi}{3}) = \frac{\sqrt {3}}{2}[/math]

[math]sin(\frac{\pi}{6}) = \frac{1}{2}[/math]

[math]sin(\frac{\pi}{2}) = 1[/math]

[math]cos(\frac{\pi}{3}) = \frac{1}{2}[/math]

[math]cos(\frac{\pi}{6}) = \frac{\sqrt {3}}{2}[/math]

[math]cos(\frac{\pi}{2}) = 0[/math]

[math]sin(\frac{\pi}{3}) = \frac{\sqrt {3}}{2}[/math]

 

so

 

[math]\sin (\frac{\pi}{3}) \sin (\frac{\pi}{6}) = \frac{\sqrt 3}{2} \times \frac{1}{2}

= \frac{\sqrt 3}{4} [/math]

 

which does not equal

 

[math] \cos ((\frac{\pi}{3}) - (\frac{\pi}{6})) - \cos((\frac{\pi}{3}) + (\frac{\pi}{6}))[/math]

 

which equals

 

[math] \cos(\frac{\pi}{6}) - \cos(\frac{\pi}{2}) = \frac{\sqrt {3}}{2} - 0[/math]

 

ending with

 

[math] \frac{\sqrt {3}}{2}[/math]

 

So either I've done something wrong in there, or the rule is wrong. They should equal the same thing, so I think the rule is wrong.

Phew, that was a lot of work to type that many equations through the system...

[Bignose]

at the end there, you forgot your factor of 1/2 which would then correctly make the last expression [math]\frac{\sqrt{3}}{4}[/math]

[D'Nalor]

heh heh... oops... I did forget about that one didn't I...

 

Yes, that does work. sorry...

 

Heh, I even forgot about the Cap'n's other rule... that's why I still couldn't get it right. I should just sub those other values in to the continued formula. continueing from there...

 

[math]

\frac{1}{2}(\int cos(-2x)cos(3x)dx -\int cos(6x)cos(3x)dx)

[/math]

 

[math]\frac{1}{2}(\int (\frac{1}{2}(\cos(-5x)-\cos(x)))dx - \int (\frac{1}{2}(\cos(3x)-\cos(9x)))[/math]

 

[math]\frac{1}{4}(\frac{-1}{5}\sin(-5x)-\sin(x)-\frac{1}{3}\sin(3x)+\frac{1}{9}\sin(9x))+c[/math]

 

and I could remove that quater at the front, but that seems unnesscessary at the moment. There we are, I can do the question now. thanks!

 

I looked back over the rule, and, stupid stupid, I forgot that

 

[math]\cos(x)=\cos(-x)[/math]

 

That would be why I thought that it wouldn't work. sorry...

[ydoaPs]

My bad, lack of sleep makes me read poorly.

[D'Nalor]

That's fine. It happens. Have an exam on this subject tomorrow, and thanks to you lot helping me with this, I think I should do quite well. It was the only thing that I wasn't quite sure of.

 

Thanks!

[the tree]

Well then good luck with your exam.

[D'Nalor]

Just finished my exam. Guess what? that question wasn't in there. what a suprise, and waste of time. I think I did well though. there was one hard question that I couldn't get though, and it would be nice if someone could show me how to do it:

 

show that [math]\int \frac{\sqrt{a}}{(x+a)\sqrt{x}} dx = arcsin(\frac{x-a}{x+a})[/math]

 

hence find [math]\int \frac{dx}{(4x+3)\sqrt{x}}[/math]

 

I found most of the other questions very easy, but this one was just ridiculous. help please?

[shyvera]

show that [math]\int \frac{\sqrt{a}}{(x+a)\sqrt{x}} dx = \arcsin\left(\frac{x-a}{x+a}\right)[/math]

 

 

Just differentiate the right-hand side.

[D'Nalor]

Umm... how? only rule I know for this is [math]\int \frac{dx}{\sqrt{a \times a - x \times x}} = \arcsin(\frac{x}{a})+c[/math]

This is the only rule For deriving that I can think of, and the teacher has said that it is possible to do it using this rule. but how?

(Couldn't figure out how to do powers using system)

[shyvera]

Why would you insist on integrating when you could just differentiate? The question doesn’t specifically ask you to integrate; it merely says:

 

Show that [math]\int\mbox{LHS d}x\ =\ \mbox{RHS}.[/math]

 

So, if you can show that [math]\frac{\mbox d}{\mbox dx}(\mbox{RHS})\ =\ \mbox{LHS},[/math] you have answered the question.

Edited June 18, 2010 by shyvera

[DJBruce]

Why would you insist on integrating when you could just differentiate? The question doesn’t specifically ask you to integrate; it merely says:

 

Show that [math]\int\mbox{LHS d}x\ =\ \mbox{RHS}.[/math]

 

So, if you can show that [math]\frac{\mbox d}{\mbox dx}(\mbox{RHS})\ =\ \mbox{LHS},[/math] you have answered the question.

 

I agree that this method should work, but the answer is not that apparent:

 

[math]\frac{d}{dx}\int\frac{\sqrt{a}}{(x+a)\sqrt{x}}dx= \frac{d}{dx}arsin\left(\frac{x-a}{x+a}\right)[/math]

 

[math]\frac{d}{dx}arsin\left(\frac{x-a}{x+a}\right)=\frac{\frac{1}{x+a}-\frac{x-a}{(x+a)^{2}}}{\sqrt{1-\frac{(x-a)^{2}}{(x+a)^{2}}}}[/math]

 

[math]\frac{\frac{1}{x+a}-\frac{x-a}{(x+a)^{2}}}{\sqrt{1-\frac{(x-a)^{2}}{(x+a)^{2}}}}=\frac{\sqrt{a}}{(x+a)\sqrt{x}}???[/math]

 

I did also take the antiderivative:

 

[math]\int\frac{\sqrt{a}}{(x+a)\sqrt{x}}dx= 2arctan\left(\frac{\sqrt{x}}{\sqrt{a}}\right)+c[/math]

 

However this did not seem to help much. I'm guessing their is some identity involved in this that I cannot for the life of me think of.

[D'Nalor]

I spoted that you could just differentiate it, the only trouble was that I was trying to get it into the form of the rule(which obviously you can't[although in retrospect i should have noticed that]). I really should have noticed that I could have just done

[math]

\frac{d}{dx}arsin\left(\frac{x-a}{x+a}\right)=\frac{\frac{1}{x+a}-\frac{x-a}{(x+a)^{2}}}{\sqrt{1-\frac{(x-a)^{2}}{(x+a)^{2}}}}

[/math]

Which would have been much easier than what I was trying to do... now at least i understand how I was supposed to have done it.

There might have been some identity, but It mightn't have been one that is very common or easy.

[DJBruce]

I am glad you save you could use FTOC Pt. I to go about this a different way, but were you able to show:

[math]

\frac{\frac{1}{x+a}-\frac{x-a}{(x+a)^{2}}}{\sqrt{1-\frac{(x-a)^{2}}{(x+a)^{2}}}}=\frac{\sqrt{a}}{(x+a)\sqrt{x}}???

[/math]

 

because to successfully prove that

 

[math]

\int \frac{\sqrt{a}}{(x+a)\sqrt{x}} dx = arcsin(\frac{x-a}{x+a})

[/math]

 

You would need to show that those are equal, which at least I wasn't able to do.

[shyvera]

[math]\frac{\mathrm d}{\mathrm dx}\left[\arcsin\left(\frac{x-a}{x+a}\right)\right][/math]

 

[math]=\ \frac{\mathrm d}{\mathrm dx}\left[\arcsin\left(1-\frac{2a}{x+a}\right)\right][/math]

 

[math]=\ \frac1{\sqrt{1-\left(1-\frac{2a}{x+a}\right)^2}}\cdot\frac{2a}{(x+a)^2}[/math]

 

[math]=\ \frac{2a}{(x+a)^2\sqrt{\frac{4a}{x+a}-\frac{4a^2}{(x+a)^2}}}[/math]

 

[math]=\ \frac{2a}{(x+a)^2\cdot\frac{\sqrt{4a(x+a)-4a^2}}{x+a}}[/math]

 

[math]=\ \frac{2a}{(x+a)\sqrt{4ax}}[/math]

 

[math]=\ \frac a{(x+a)\sqrt{ax}}[/math]

 

[math]=\ \frac{\sqrt a\sqrt a}{(x+a)\sqrt a\sqrt x}[/math]

 

[math]=\ \frac{\sqrt a}{(x+a)\sqrt x}[/math]

Edited June 19, 2010 by shyvera

[DJBruce]

Thanks for the solution. The first line was the thing, which I was missing.

[D'Nalor]

Oops, silly me, I got the first part but not the second... I really should have tried just doing that second part as well. It flows quite nicely after that. oh well, at least it was the only one that I couldn't get. thanks for that. nothing else I need to ask at the moment, and it will be unlikely that I'll need to again on this topic for quite a while.

 

Thanks again for your help!