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Tue 14 Aug, 2007 01:41 pm
Please help me with this Integral problem. How does one integrate a function of the following form:
(Asinx+Bcosx)/(Csinx+Dcosx)
For example, integrating
(3cosx - 4sinx)/(3sinx + 4cosx)
A hint given is that you could add or multiply to give a simpler integral. I think that hint is dodgy and meant to throw you of track (yes, they do that nowadays...)
Here is a possible solution, is it correct?
(Asinx+Bcosx)/(Csinx+Dcosx) =
A(C^2+D^2)sinx+ B(C^2+D^2)cosx)/(C^2+D^2)(Csinx+Dcosx)=
(one, two, skip a few, 99, 100:)
...=
[(AC+BD)(Csinx+Dcosx)+(BC-AD)(Ccosx-Dsinx)]/[(C^2+D^2)(Csinx+Dcosx)]
Now separate:
= (AC+BD)/(C^2+D^2) + [(BC-AD)/(C^2+D^2)]*[(Ccosx-Dsinx)/(Csinx+Dcosx)]
The integral of the first part is (AC+BC)x/(C^2+D^2) since it is just a constant and using substitution
u = Csinx+Dcosx the second part becomes
(BC-AD)/(C^2+D^2)*
ln|Csinx+Dcosx|
Which seems a very contrived way of doing things; is there not a simpler way? Is that stuff even correct??? Please help, this problem is driving me up the wall!
Suggestion
Try using the substitution t=tan(1/2)x
whence sinx=2t/1+t^2 & cosx=(1-t^2)/(1+t^2)
This should give a quotient of quadratics in t such that
INT f(x).dx can be rewritten INT g(t).2dt/(1+t^2)
I think the next move is to get partial fractions but its many years since I did this stuff.
Thanks fresco. I just discovered (with a bit of help ofcourse) how to solve these thingys.
You have to get the numerator in terms of real multiples of the denominator or its derivative, then you can simplify and it becomes simple.
(Asinx+Bcosx)/(Csinx+Dcosx)
We want
Asinx + Bcosx = p(Csinx + Dcosx) + q[(Csinx + Dcosx)']
Asinx + Bcosx = pCsinx+ pDcosx + qCcosx - qDsinx
Therefore:
A = pC - qD
B = pD + qC
Solve for p and q:
q=(BC-AD)/(D^2 + C^2)
p = (A(D^2+C^2) + D(BC-AD))/(C(D^2+C^2))
So the numerator can be written as:
Asinx + Bcosx = p(Csinx+ Dcosx) + q(Ccosx - Dsinx)
using p and q above. So (Asinx+Bcosx)/(Csinx+Dcosx) (what we want to integrate) becomes
[p(Csinx+ Dcosx) + q(Ccosx - Dsinx)]/(Csinx+Dcosx)
= p + q(Csinx+Dcosx)'/(Csinx + Dcosx)
which can easily be solved by substitution. It is essentially the same thing done in the first post, but now it makes sense.
Also, how does one integrate:
(x^2).(x+3)^(1/3) ?
Thanks.
That looks like "integration by parts" using
INT u.(dv/dx).dx=uv - INT v.(du/dx).dx
Try using u=x^2 and dv/dx = (x+3)^1/3
[whence v=(3/4)(x+3)^4/3]
You might have to apply the method again to the blue term which is still a product of functions of x
Thanks for the suggestion fresco, I didn't think of using integration by parts twice, even though I've done some telescoping integrals.
I preffer the notation:
INT u.v' = uv - INT u'. v
but that's trivial...
It's just a bit tricky deciding which rule you're going to use, but it comes with practise.
Thanks a lot fresco.
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