Ducky Havok Posted April 6, 2005 Share Posted April 6, 2005 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. Link to comment Share on other sites More sharing options...
Dave Posted April 6, 2005 Share Posted April 6, 2005 Do you mean indefinite integral? Link to comment Share on other sites More sharing options...
Ducky Havok Posted April 6, 2005 Author Share Posted April 6, 2005 Yes, and now I feel stupid for messing that up twice in the same post I'll go back and fix that though Link to comment Share on other sites More sharing options...
Dave Posted April 6, 2005 Share Posted April 6, 2005 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. Link to comment Share on other sites More sharing options...
Dave Posted April 6, 2005 Share Posted April 6, 2005 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]. Link to comment Share on other sites More sharing options...
Ducky Havok Posted April 7, 2005 Author Share Posted April 7, 2005 One small thing, in my problem the bottom part is squared. [math]\int \frac{1}{{(a\sin(x) + b\cos(x))^2}} \, dx[/math] Link to comment Share on other sites More sharing options...
Dave Posted April 7, 2005 Share Posted April 7, 2005 Indeed; I forgot to put that in Answers are still right though; edited the above posts to make sure that it makes sense now. Link to comment Share on other sites More sharing options...
Ducky Havok Posted April 7, 2005 Author Share Posted April 7, 2005 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. Link to comment Share on other sites More sharing options...
Dave Posted April 7, 2005 Share Posted April 7, 2005 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. Link to comment Share on other sites More sharing options...
Ducky Havok Posted April 7, 2005 Author Share Posted April 7, 2005 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... Link to comment Share on other sites More sharing options...
bishnu Posted April 9, 2005 Share Posted April 9, 2005 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 Link to comment Share on other sites More sharing options...
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