WHAT NOW?

>>8450404
Chances are, there's no analytical solution to this.

It means no function can describe it. If you really want an anti derivative solution you need to define it and study it's properties.

You can use Riemann sums to an arbitrary number( use a shitton for better accuracy)
or put it into mathematica or whatever program you wan to use.

>>8450406
That's fucked up because it's on my textbook and you're supposed to do it with U Substitution. And no, they do not give away the answer.

>>8450421
I couldn't crack it. I could be possible that it's an error with the textbook, but maybe we both just suck at basic calculus?

you sure you copied that one right? Cot instead of Tan or Cos instead of Sin and it would be peaches.

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>>8450441
Straight from it. I don't even know anymore. I checked more recent versions and they haven't changed it.

>>8450404
Use Implicit Integration.
>>8450433
>>8450421
>>8450406
>>8450413

you all suck

>>8450450
OP said U substitution, so i tried that.

>>8450404

That shit is not even differentiable, it's asymptotic at pi/2+pi*n

>>8450450
>>8450453
>>8450404
No, you can u-sub it. I'm surprised Wolfram can't do it, but it's definitely possible

>>8450453
shut up brainlet,

Anyways OP if you're still struggling use this
[eqn] \mbox{Ei}(x) = \int_{-\infty}^x dt~\frac{e^u}{u} [/eqn]

>>8450473
dt should be du

>>8450464
That doesn't mean it's never differentiable - only that it isn't at those points.

>>8450467
>>8450404
Bet the NSpire CAS will do it. Better engine than Wolfram.

u = sin(x)
du = cos(x) dx

[eqn] \int \frac{\tan(x)}{\log(\sin(x))} dx = \int \frac{\sin(x) \cos(x)}{(1 - \sin^2(x))\log(\sin(x))} dx = \int \frac{u}{(1 - u^2) \log(u)} du[/eqn]
The rest is trivial.

>>8450762
fuking kek'd

>>8450762

Then do the rest

>>8450762
Do it

u=1/ln(sinx)
du=dx/(cosx/sinx)=dx/cotx=tanxdx

Then it's just the integral of du/u.

>>8450948
More like
[eqn] u = \frac{1}{\log(\sin(x))} [/eqn]
[eqn] du = \frac{-1}{\left(\log(\sin(x)) \right)^2} \frac{1}{\sin(x)} \cos(x) dx = \frac{- \cot(x)}{\left(\log(\sin(x)) \right)^2} dx[/eqn]

You have to learn the chain rule.

>>8450404
Ask it on stackexchange. Cleo will provide an answer.

>>8450445
Wtf is sen(x)? And where is the infinitesimal?

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i got it to pic related.
can it be solved?

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>>8451134
Nope.

>>8451146
bummer

>>8451146
you could try expanding 1/(1-x) into the sum of all powers of x

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>>8450404
If only we could see a sign.

>>8450404
Wolfram isn't giving you an answer because the indefinite integral of log(sinx) can't be expressed with elementary functions. Try redefining the entire thing as a hyperbolic trig and then take the anti derivative it might be possible then.

http://reference.wolfram.com/legacy/v5/TheMathematicaBook/AdvancedMathematicsInMathematica/Calculus/3.5.7.html

>>8450971
It appears to be Spanish sine.

It's a lie. Can't be done.

Many integrals cannot be evaluated on W|A in terms of elementary functions. An example is the integral of ((x^2 - 1)/(x^2 + 1)) * (1/sqrt(x^4 + 1)) dx
I'll leave this as an exercise. (Should be easy.)

>>8450413
>and study it is properties

>>8450404
>W-what now, /sci/?
Stab ur fucking prof. Now. Not even fucking kidding.

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>>8450404
holy shit sci

>>8450404
I thought I found a closed form solution using li(x) and W(x) but i made an error.

idk anon, go to stack exchange and ask cleo, tough one. There is 0% this isn't a typo.

substitute [math] u = sin(x) [/math] then [math] du = cos(x). [/math]
leaving the integral as:
[eqn] \int \frac{u* du}{(1-u^2)*ln(u)} [/eqn]
since [math] u = sin(x) [/math] we can assume quite safely that [math] sin(x) < 1 [/math] and then make the series substitution
[eqn] \frac{1}{1-u^2} = \sum_{k=0}^{\infty} u^{2k} [/eqn]
this in turn leaves out integral as:
[eqn] \int \sum_{k=0}^{\infty} \frac{u^{2k+1} (u-1)^{k} (-1)^{k+1}}{k} [/eqn]
which mathematica tells me has beta function solutions

>>8452468
shoot, i mean [math] sin(x) = u \implies u < 1 [/math]

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I did it. Now where is my 300k starting salary?

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>>8452516
You did it wrong.

>>8452516
Kek

>>8450421
Do a U sub and you'll get, u/log (u), then integrate by parts.

>>8450464
Functions with asymptotes can still be differentiated and integrated, just consider y=1/x.
Also, even if it wasn't differentiable, it might still be able to be integrated.