WHAT NOW?

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

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]

dt should be du

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

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.

fuking kek'd

Then do the rest

Do it

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

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

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.

Ask it on stackexchange. Cleo will provide an answer.

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

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

Nope.

bummer

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

If only we could see a sign.

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.

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

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.)

>and study it is properties

>W-what now, Veeky Forums?
Stab ur fucking prof. Now. Not even fucking kidding.

holy shit sci

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

idk user, 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

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

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

You did it wrong.

Kek

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

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.