Integral Problem

[Ducky Havok]

How would you solve the following problem?

 

[math]\int \frac{1}{{(asin{x}+bcos{x})^2}}, dx[/math] with a and b being constants.

 

I'm pretty sure the answer is [math]\frac{2tan{\frac{x}{2}}}{(-btan{\frac{x}{2}}+\sqrt{2ab+1}+a)(btan{\frac{x}{2}}+\sqrt{2ab+1}-a)}+C[/math], but I want to know how to do it.

[Dave]

Do you mean indefinite integral?

[Ducky Havok]

Yes, and now I feel stupid for messing that up twice in the same post I'll go back and fix that though

[Dave]

I'm having some problems with this solution. I'm usually fairly good with the integrals, but it's been a while. Mathematica gives this answer:

 

[math]\int \frac{1}{(a\sin x + b\cos x)^2} \, dx = \frac{\sin x}{b(b\cos x + a\sin x)}[/math]

 

Excluding the constant of integration, of course. However, I can't get this answer by any means I've tried, which is annoying. If you differentiate it and simplify it down, it definately works. Hmm.

[Dave]

Update: I now have it down to:

 

[math]\int \frac{1}{(a\sin x + b\cos x)^2} \, dx = \frac{b\sin x- a\cos x}{(a^2 + b^2)(b\cos x + a\sin x)}[/math]

 

I'll be damned if I can simplify it down though

 

Hint: Use the fact that [math]a\sin x + b\cos x \equiv R\cos(x-\phi)[/math].

[Ducky Havok]

One small thing, in my problem the bottom part is squared.

[math]\int \frac{1}{{(a\sin(x) + b\cos(x))^2}} \, dx[/math]

[Dave]

Indeed; I forgot to put that in

 

Answers are still right though; edited the above posts to make sure that it makes sense now.

[Ducky Havok]

Well, on the bright side, my first answer and that answer are the exact same, that one is just super simplified compared to mine. On the dim side (because if there's a bright side there has to be a dim), I still don't get how to do it even with your hint. What is [math]R[/math] and [math]\phi[/math]? At first I thought [math]\phi[/math] was just theta but I put my mouse over it and it said phi. I haven't seen either of those symbols yet, but I'm still in baby calc. One of my friends just asked me this question because they couldn't get it.

[Dave]

Basically, [math]R[/math] and [math]\phi[/math] are just constants which satisfy:

 

[math]R^2 = a^2 + b^2[/math], [math]\tan\phi = \tfrac{b}{a}[/math]. If you expand the RHS you'll see that it's equal to the LHS.

[Ducky Havok]

Aren't those the conversions to polar coordinates? I'm just going off memory, we haven't gotten to those yet. So do you get something along the lines of

[math]-\frac{\sqrt{R^2-U^2}}{{R^2}U}[/math] with [math]U=RCos(x-\phi)[/math] ? I think I confused myself worse there actually...

[bishnu]

okay the intergral equals 1/(a^2tan(x)+b) heres how i got it btw i hate latex so just try to follow along

 

okay factor out a cos from the bottom 1/cos^2(atan+b)^2

 

which equals sec^2/(atan+b)^2

 

then set u=atan+b

so then the intergral becomes 1/au^2 now intergrate and get -1/au then subsitute back in and get -1/(a^2tanx+ba)+c