Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Is my input below correct? I've received no response from Mathematica.

N[Limit[Integrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, Pi/2}], n -> Infinity]]

Thank you in advance for your feedback!

share|improve this question
    
The answer (output) should be $\pi/4$. –  Chris's sis Feb 20 '13 at 9:32
    
The result of the integral seems to be Pi/4 for any n. –  b.gatessucks Feb 20 '13 at 10:39
    
@b.gatessucks: Right. This is straightforward by letting $x=\pi/2-y$. –  Chris's sis Feb 20 '13 at 10:52
1  
so problem solved ? –  b.gatessucks Feb 20 '13 at 11:03
    
@b.gatessucks: I'm trying to learn to deal with such things by using Mathematica. –  Chris's sis Feb 20 '13 at 11:07
show 3 more comments

1 Answer

up vote 4 down vote accepted

Not an answer, just will not fit in the comment.

Mathematica can't seem to be able to integrate Sin[x]^n/(Sin[x]^n + Cos[x]^n analytically. I tried using the formulas for Sin[x]^n and Cos[x]^n for odd n with the hope it will help. But still no luck. Using Wiki, the formulas are from here

mysin[x_,n_] := (2/(2^n)) Sum[(-1)^((n - 1)/2 - k) Binomial[n, k]*
     Sin[(n - 2 k) x], {k, 0, (n - 1)/2}];

mycos[x_,n_] := (2/(2^n)) Sum[Binomial[n, k]*Cos[(n - 2 k) x], {k, 0, (n - 1)/2}];

integrand = 
 Assuming[Element[n, Integers] && n > 0 && Element[x, Reals], 
  FullSimplify[mysin[x, n]/(mysin[x, n] + mycos[x, n])]];

res = ComplexExpand[integrand];
Integrate[res, {x, 0, Pi/2}]

Its been running for 30 minutes.

But doing numerical integration

Table[
 NIntegrate[Sin[x]^n/(Sin[x]^n + Cos[x]^n), {x, 0, Pi/2}], {n, 0, 10}]

(*Out[1]= {0.785398, 0.785398, 0.785398, 0.785398, 0.785398, 0.785398, \
0.785398, 0.785398, 0.785398, 0.785398, 0.785398}*)

So may be you do not not need to do analytical integration after all? Are you sure this integrand can be integrated analytically?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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