3
$\begingroup$

I am trying to understand how WolframAlpha reduces the following trig expression $$ \frac{\ln \left(\sin \left(\frac{\alpha x}{2}\right)+\cos \left(\frac{\alpha x}{2}\right)\right)}{\alpha }-\frac{\ln \left(\cos \left(\frac{\alpha x}{2}\right)-\sin \left(\frac{\alpha x}{2}\right)\right)}{\alpha }$$ into $$\frac{\ln (\tan (\alpha x)+\sec (\alpha x))}{\alpha }$$

I tried using TrigExpand, TrigReduce and FullSimplify with no use.

CODE:

FullSimplify[Integrate[Sec[\[Alpha]*x], x], Assumptions -> {\[Alpha] > 0 && Element[x, Reals]}]
$\endgroup$
  • $\begingroup$ Did you try Assumptions->[Alpha]>0&& x [Element] Reals? $\endgroup$ – user64494 Jan 19 '18 at 18:23
  • $\begingroup$ @user64494 yes I’ll post the code. It doesn’t work $\endgroup$ – DMH16 Jan 19 '18 at 18:27
  • $\begingroup$ The command with the option Assumptions -> [Alpha] > 0 && x >= 0 && x <= Pi/(4*[Alpha]) performs (2 ArcTanh[Tan[(x [Alpha])/2]])/[Alpha]. $\endgroup$ – user64494 Jan 19 '18 at 18:36
  • $\begingroup$ This will get you part way there Simplify[Log[Sin[a x/2]+Cos[a x/2]]/a-Log[Cos[a x/2]-Sin[a x/2]]/a, Sin[a x/2]+Cos[a x/2]>0 && Cos[a x/2]-Sin[a x/2]>0]To finish you probably need to think more about zeros in denominators $\endgroup$ – Bill Jan 19 '18 at 18:39
  • 3
    $\begingroup$ Note that Integrate[Sec[α x], x] /. {x -> 1, α -> 3.} is different from Log[Tan[α x] + Sec[α x]]/α /. {x -> 1, α -> 3.}, so the functions are not the same. If that's not important, maybe FullSimplify[Integrate[Sec[α x], x], TransformationFunctions -> {Automatic, Log@*Exp}] suffice? $\endgroup$ – Coolwater Jan 19 '18 at 20:25
2
$\begingroup$

May be this can be a start. I can't find method for final transformation yet. May be it needs special rule.

ClearAll[x,a,A0,B0];
rep    =  {A0->Sin[a x/2],B0->Cos[a x/2]}
expr   =  Log[A0+B0]-Log[B0-A0] ;
result =  (FullSimplify[Log[Exp[expr]]]/.rep)/a

$$ \frac{\log \left(\frac{\sin \left(\frac{a x}{2}\right)+\cos \left(\frac{a x}{2}\right)}{\cos \left(\frac{a x}{2}\right)-\sin \left(\frac{a x}{2}\right)}\right)}{a} $$

$\endgroup$
  • $\begingroup$ Interesting! Thank you for the answer. I won't accept one yet, since I want to see if other ideas come by. $\endgroup$ – user372003 Jan 19 '18 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.