- #1

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I'm told that this integration can't be solved, can it ?

ʃ dx/(sin (x) . exp((csc x)^2)

ʃ dx/(sin (x) . exp((csc x)^2)

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- Thread starter custer
- Start date

- #1

- 13

- 0

I'm told that this integration can't be solved, can it ?

ʃ dx/(sin (x) . exp((csc x)^2)

ʃ dx/(sin (x) . exp((csc x)^2)

- #2

HallsofIvy

Science Advisor

Homework Helper

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- #3

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I'm told that this integration can't be solved, can it ?

ʃ dx/(sin (x) . exp((csc x)^2)

Make it so. Suppose that was all you were told. What would you do? Numerically counts as "solved" in my book. How about a power series? Can I use that (in principle) and integrate over it's radius of convergence? Ain't that a "solution"? Find a way . . . try. That's what math is about. :)

- #4

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Make it so. Suppose that was all you were told. What would you do? Numerically counts as "solved" in my book. How about a power series? Can I use that (in principle) and integrate over it's radius of convergence? Ain't that a "solution"? Find a way . . . try. That's what math is about. :)

And if all else fails, and it's a useful enough integral, then give it a name, e.g.

"For all real x, we define

[tex]cus(x) = \int_{-\infty}^x \frac{dx}{\sin(x) \exp((\csc x)^2)}[/tex]"

Presto, solved! Now all you need to do is lobby for a calculator button. Oh, and maybe establish some facts about the function. If people like it enough, they might assign your name to it. It worked for Bessel.

- #5

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[tex]f(x)=\frac{1}{\sin(x)e^{\sec^2(x)}}[/tex]

then I'd suggest we define:

[tex]\text{cus}(x)=\int_0^x \frac{1}{\sin(x)e^{\sec^2(x)}}[/tex]

and I'd like to establish the first property of this [tex]\text{cus}(x)[/tex] (why? just be-cus):

[tex]\text{cus}(-x)=-\text{cus}(x)[/tex]

Also, just to get some empirical data about [tex]\text{cus}[/tex] (why, just . . . ok, I'll stop it), I calculated the first 25 terms of the Taylor series about the point [tex]x=\pi/2[/tex] and then calculated the antiderivative of those terms, substituted the limits [tex]\pi/4[/tex] and [tex]3\pi/4[/tex] then compared it to a direct numerical integration of the integrand. The results were accurate to 6 places:

Code:

```
f[x_] := 1/(Sin[x]*Exp[Csc[x]^2]);
tay[x_] := Normal[Series[f[x],
{x, Pi/2, 25}]]
cus[x_] = Integrate[tay[x], x]
N[cus[3*(Pi/4)] - cus[Pi/4]]
NIntegrate[f[x], {x, Pi/4, 3*(Pi/4)}]
```

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