Find the integral:
$$
I=\int\frac{1}{\sin\sqrt{x}}dx
$$
Well, here is what I've done so far:
$$
\begin{aligned}
&\sqrt{x}=t\Rightarrow dt=\frac{dx}{2\sqrt{x}},\ \ dx=2tdt\\
&I=2\int\frac{t}{\sin t}dt=\ ?
\end{aligned}
$$
The problem is to calculate the last integral. $t$ in the numerator makes it rather insolvable.
Best Answer
Hint:
Recall that $$\sin\theta = \frac{\exp(\mathrm i\theta) - \exp(-\mathrm i\theta)}{2\mathrm i}$$
Therefore, $$\int\frac t{\sin t}\,\mathrm dt = \int\frac{2\mathrm it}{\exp(\mathrm it) - \exp(-\mathrm it)}\,\mathrm dt = 2\mathrm i\int\frac{t\exp(\mathrm it)}{\exp(\mathrm 2\mathrm it)-1}\,\mathrm dt$$
Let $u = \mathrm it\implies\mathrm dt = -\mathrm i\mathrm du$.
$$2\int\frac{\mathrm it\exp(\mathrm it)}{\exp(\mathrm 2\mathrm it)-1}\,\mathrm dt\equiv -2\int\frac{u\exp(u)}{\exp(2u) - 1}\,\mathrm du$$
Let $v = \exp(u)\implies \mathrm du = \exp(-u)\mathrm dv$. Then, $u = \ln(v)$ and $\exp(2u) = v^2$.
$$\int\frac{u\exp(u)}{\exp(2u) - 1}\,\mathrm du\equiv\int\frac{\ln(v)}{v^2 - 1}\,\mathrm dv$$