Let $k$ and $x$ be real numbers. Let:
$$I= \int \frac{dx}{\sin x \sqrt{k^2+\sin^2 x}}$$
Let $u=\sin x$, then $du=\sqrt{1-u^2}dx$. Then
$$I= \int \frac{du}{u\sqrt{k^2+u^2}\sqrt{1-u^2}}$$
To go further any recommendations? Not sure if $u=k\tan y$ may help.
A mysterious solution. We apply the substitution
$$ t = \frac{k\cos x}{\sqrt{k^2+\sin^2 x}}. $$
This substitution has the property that
$$ \color{blue}{1 - t^2 = \frac{(k^2+1)\sin^2 x}{k^2+\sin^2 x}} \qquad\text{and}\qquad \color{red}{\mathrm{d}t = -\frac{k (k^2+1) \sin x}{(k^2+\sin^2 x)^{3/2}} \, \mathrm{d}x}. $$
Hence it follows that
\begin{align*} I &= \int \frac{\mathrm{d}x}{\sin x \sqrt{k^2 + \sin^2 x}} \\ &= \int \frac{1}{\sin x \sqrt{k^2 + \sin^2 x}} \color{red}{\biggl( -\frac{(k^2+\sin^2 x)^{3/2}}{k (k^2+1) \sin x} \, \mathrm{d}t \biggr)} \\ &= - \int \frac{\color{blue}{k^2+\sin^2 x}}{k \color{blue}{(k^2+1) \sin^2 x}} \, \mathrm{d}t \\ &= -\frac{1}{k} \int \frac{\mathrm{d}t}{1 - t^2}. \end{align*}
The last integral can be easily computed, yielding
$$ I = -\frac{1}{k} \operatorname{artanh}(t) + \mathsf{C} = \bbox[padding:8px;border:1px navy dotted; color:navy;]{ - \frac{1}{k} \operatorname{artanh}\biggl(\frac{k \cos x}{\sqrt{k^2 + \sin^2 x}}\biggr) + \mathsf{C} }. $$
A less mysterious solution. Let $p = \sqrt{k^2 + 1}$ for simplicity. Then
\begin{align*} I &= \int \frac{\sin x}{(1-\cos^2 x)\sqrt{p^2-\cos^2 x}} \, \mathrm{d}x \\ &= - \int \frac{1}{(1-u^2)\sqrt{p^2-u^2}} \, \mathrm{d}u \tag{$u=\cos x$} \\ &= - \int \frac{\cosh\phi}{1 - k^2\sinh^2 \phi} \, \mathrm{d}\phi \tag{$u=p\tanh\phi$} \\ &= - \frac{1}{k} \operatorname{artanh}(k \sinh\phi) + \mathsf{C} \\ &= - \frac{1}{k} \operatorname{artanh}\biggl(\frac{k u}{\sqrt{p^2 - u^2}}\biggr) + \mathsf{C} \\ &= - \frac{1}{k} \operatorname{artanh}\biggl(\frac{k \cos x}{\sqrt{k^2 + \sin^2 x}}\biggr) + \mathsf{C}. \end{align*}
Hint Rewrite the integral as $$\int \frac{\sin x}{(1 - \cos^2 x) \sqrt{k^2 + 1 - \cos^2 x}} \,dx,$$ which suggests the substitution $$\cos x = \sqrt{k^2 + 1} \sin \theta, \qquad - \sin x \,dx = \sqrt{k^2 + 1} \cos \theta \,d\theta ,$$ which in turn yields an integrand rational in $\sin \theta, \cos \theta$: $$\int \frac{d\theta}{k^2 - (1 + k^2) \cos^2 \theta} .$$ Now, the substitution $t = \tan \theta$ rationalizes the integral: $$\int \frac{dt}{k^2 t^2 - 1} .$$
$$- \frac1k \operatorname{artanh} k t + C = \boxed{-\frac1k \operatorname{artanh} \frac{k \cos \theta}{k^2 + \sin^2 \theta} + C}$$
Multiplying the numerator and denominator by $\sec^2x$ and letting $t=\tan x$ yields $$ I=\int \frac{\sec ^2 x d x}{\tan x \sqrt{k^2 \sec ^2 x+\tan ^2 x}}= \int \frac{d t}{t \sqrt{k^2+\left(k^2+1\right) t^2}}= \int \frac{td t}{t^2 \sqrt{k^2+\left(k^2+1\right) t^2}} $$ $\text { Let } \quad y=\sqrt{k^2+\left(k^2+1\right) t^2}$, then $$ \begin{aligned} I & =\int \frac{d y}{y^2-k^2} \\ & =\frac{1}{2 k} \ln \left|\frac{y-k}{y+k}\right|+C \\ & =\frac{1}{2 k} \ln \left|\frac{\sqrt{k^2+\left(k^2+1\right) t^2}-k}{\sqrt{k^2+\left(k^2+1\right) t^2}+k}\right|+C\\ & =\frac{1}{2 k} \ln \left|\frac{\sqrt{k^2 \sec ^2 x+\tan ^2 x}-k}{\sqrt{k^2 \sec ^2 x+\tan ^2 x+k}}\right|+C \\ & =\frac{1}{2 k} \ln \left|\frac{\sqrt{k^2+\sin ^2 x}-k\cos x}{\sqrt{k^2+\sin ^2 x}+k \cos x}\right|+C \end{aligned} $$ Wish it helps.
I might be incorrect and it's been a while since I did u-substitution. But you should combine the square roots: $$I=\int\dfrac{du}{u\sqrt{k^2-k^2u^2+u^2-u^4}}$$Then you can replace $u$ with $k\tan(y)$ and further simplify it.
