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.

Do you mean indefinite integral?

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

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.

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].

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

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

Indeed; I forgot to put that in

 

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

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.

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.

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...

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