Find $\int\frac{1}{\sin\sqrt{x}}dx$

Question

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.

Answer 1

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 $\exp(2u) = v^2$.

$$\int\frac{u\exp(u)}{\exp(2u) - 1}\,\mathrm du\equiv\int\frac{\ln(v)}{v^2 - 1}\,\mathrm dv$$

Answer 2

Starting from @an4s's answer $$\int\frac{\log(v)}{v^2 - 1}\, dv=\frac 12 \left(\int\frac{\log(v)}{v - 1}\, dv -\int\frac{\log(v)}{v + 1}\, dv\right)$$ and one integration by parts leads to $$\int\frac{\log(v)}{v + a}\, dv=\log (v) \log \left(1+\frac{v}{a}\right)+\text{Li}_2\left(-\frac{v}{a}\right)$$ So, for $a=1$ $$\int\frac{\log(v)}{v^2 - 1}\, dv=\frac 12\Big(\text{Li}_2(v)-\text{Li}_2(-v) \Big)-\log (v) \tanh ^{-1}(v)$$