A simple question, but I can't seem to find the answer. I want to calculate the definite integral of the function $f(x)=x\cdot\csc(x)$ from $0$ to $1$:
$$\int_0^1 x\csc(x)\, dx. $$
And from there, calculate the same integral, but with $x^3\csc(x)$, $x^4\csc(x)$, and $x^6\csc(x)$.
I'm almost convinced it's impossible, but my professor suggested to use the change of variable $u=\tan(\frac{x}{2})$. Is there any way to calculate these integrals?
One can evaluate this in the following way: $$\int _0^1\frac{x}{\sin \left(x\right)}\:dx$$ By using Weierstrass sub we get: $$\int _0^{\tan \left(\frac{1}{2}\right)}\frac{2\arctan \left(t\right)}{\frac{2}{1+t^2}}\:\frac{2}{t^2+1}dt=2\int _0^{\tan \left(\frac{1}{2}\right)}\frac{\arctan \left(t\right)}{t}\:dt$$ $$=2\:\text{Ti}_2\left(\tan \left(\frac{1}{2}\right)\right)\approx1.059762$$ Where i used the inverse tangent integral identity, see here for more info.