What to do when Wolfram doesn't give answer?

[Settembrini]

I'm trying to compute following integral (Wolfram doesn't give answer):
\int\sqrt{E-Bk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}-k\frac{cos(kr)}{sin(kr)}\sqrt{D+Fk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}}-\frac{Ak^{2}}{sin^{2}(kr)}}dr
where A,B,C,D,E,F,k are constants.
Substitution t=sin(kr) leads to
\int\sqrt{\frac{E}{k^{2}(1-t^{2})}-\frac{B}{t^{2}}-\frac{1}{kt\sqrt{1-t^{2}}}\sqrt{D+Fk^{2}\frac{1-t^{2}}{t^{2}}}-\frac{A}{t^{2}(1-t^{2})}}dt
The last integral is equal to
\int\frac{1}{t\sqrt{1-t^{2}}}\sqrt{(\frac{E}{k^{2}}+B)t^{2}-(A+B)-\sqrt{(F-\frac{D}{k^{2}})t^{4}+(\frac{D}{k^{2}}-2F)t^{2}+F}}dt
I don't have any idea what to do now. Maybe the substitution t=sin(kr) is not appropriate here?
Any help is appreciated.

 

[Mark44]

I'm trying to compute following integral (Wolfram doesn't give answer):
\int\sqrt{E-Bk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}-k\frac{cos(kr)}{sin(kr)}\sqrt{D+Fk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}}-\frac{Ak^{2}}{sin^{2}(kr)}}dr
where A,B,C,D,E,F,k are constants.
Substitution t=sin(kr) leads to
\int\sqrt{\frac{E}{k^{2}(1-t^{2})}-\frac{B}{t^{2}}-\frac{1}{kt\sqrt{1-t^{2}}}\sqrt{D+Fk^{2}\frac{1-t^{2}}{t^{2}}}-\frac{A}{t^{2}(1-t^{2})}}dt
The last integral is equal to
\int\frac{1}{t\sqrt{1-t^{2}}}\sqrt{(\frac{E}{k^{2}}+B)t^{2}-(A+B)-\sqrt{(F-\frac{D}{k^{2}})t^{4}+(\frac{D}{k^{2}}-2F)t^{2}+F}}dt
I don't have any idea what to do now. Maybe the substitution t=sin(kr) is not appropriate here?

It's not obvious to me that you did the substitution correctly. If t = sin(kr), then dt = kcos(kr)dr. I might be wrong, but it appears that you replaced sin(kr) by t and dr by dt.

Even if you did the substitution correctly, it might not be helpful to do.

Also, homework and homework-type problems should be posted in the Homework & Coursework sections, not in the technical math sections. I am moving your post to Calculus & Beyond section under Homework & Coursework.

 

[Settembrini]

Sorry for the wrong forum.
I think I have done substitution correctly - extracting k^{2}cos^{2}(kr) from the expression in the first integral gives:
\int kcos(kr)\sqrt{\frac{E}{k^{2}cos^{2}(kr)}-\frac{B}{sin^{2}(kr)}-\frac{1}{ksin(kr)cos(kr)}\sqrt{D+Fk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}}-\frac{A}{sin^{2}(kr)cos^{2}(kr)}}dr
So the expression kcos(kr)dr is present.

 

[Mark44]

I don't see that that is much help. Something different you might try is to rewrite this:

$$\int\sqrt{E-Bk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}-k\frac{cos(kr)}{sin(kr)}\sqrt{D+Fk^{2}\frac{cos^{2}(kr)}{sin^{2}(kr)}}-\frac{Ak^{2}}{sin^{2}(kr)}}dr $$
as this
$$\int \sqrt{E-Bk^2cot^2(kr) - k cot(kr) \sqrt{D+Fk^2cot^2(kr)-Ak^2csc^2(kr)}}dr $$

I don't guarantee that this will be helpful, either. Where I'm going with this is possibly factoring the first three terms in the outer radical, and replacing the cot²(kr) term in the inner radical by csc²(kr) - 1.

Where did this problem come from? It looks complicated enough that it might not be solvable by analytic means.