Problem with Wolfram's online integrator

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The discussion centers around the challenges of integrating the function cos(x)^x. Initial attempts using Wolfram's online integrator and other platforms like Quickmath and Axiom yielded no satisfactory results, indicating that there may not be a known closed form for this integral. Participants clarify that the confusion arises from the syntax, specifically whether the integral is of cos(x) raised to the power of x or vice versa. A suggestion is made that since cos(x) is bounded between -1 and 1, the integral of cos(x)^x might be conceptually similar to integrating other bounded functions like sin(x)^x. However, the consensus is that there is no closed form for the indefinite integral of cos(x)^x, and numerical methods may be necessary for evaluation.
BenVitale
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I tried to do integral[cos(x)]^x dx

but Wolfram's online integrator reported that it couldn't do it

Am I missing something?
 
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From your syntax it is not clear if you want to take the integral of cos(x)^x or if you want to take the integral of cos(x) and then raise that to the power of x. If it is the first then there is probably no known closed form solution for that integral.
 
Wolfram is not that good.

Try www.quickmath.com, it's better; or if you're on Linux, there're plenty of computer algebra systems.

However this time quickmath also failed...very hilarious answer.

I tried it in axiom, and the answer is bad -

\int \sp{\displaystyle x} {{{\cos <br /> \left(<br /> { \%M} <br /> \right)}<br /> \sp \%M} \ {d \%M}} <br />

yacas gives the same results (as quickmath).

sympy takes infinite time to solve (as with most complex cases)

So I conclude there's something wrong with the question itself.
 
DaleSpam said:
From your syntax it is not clear if you want to take the integral of cos(x)^x or if you want to take the integral of cos(x) and then raise that to the power of x.
.

It's taking the integral of cos(x)^x

dE_logics said:
So I conclude there's something wrong with the question itself.

Maybe, maybe not. I thought of a manipulation:

I figure that for all values of x, cos x will fall in [-1,+1].
So, taking the integral of cos(x)^x is equivalent to taking the integral of x^x over [-1,0] and [0,+1]

What do you think?
 
By that logic the integral would be the same as sin(x)^x or frac(x)^x or saw(x)^x or any other function bounded between -1 and 1.
 
DaleSpam said:
By that logic the integral would be the same as sin(x)^x or frac(x)^x or saw(x)^x or any other function bounded between -1 and 1.

Oh, yes. I see your point. I'm at loss here. What do you suggest?
 
I believe there is no closed form for this integral. You will have to evaluate it numerically.
 

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