$$\int \frac {\cos x}{\sqrt {1+\sin^2 x}} \ \mathrm{d}x$$
$u = \sin x$ seems like a reasonable place to start... but we know that that is not quite right.
$\displaystyle \int \frac {1}{\sqrt {1+u^2}} \mathrm{d}u$
When we see $1+u^2$ we should be thinking of two options.
$u = \sinh t$ or $u=\tan t.$ Ultimately, either one will work. But, many Calc $1,2$ students never see the hyperbolics. I will show both approaches.
$u = \tan t, \mathrm{d}u = \sec^2 t~ \mathrm{d}t$
$\begin{aligned} \displaystyle \implies \int \frac {\sec^2 t}{\sqrt {1+\tan^2 t}} \mathrm{d}t
&=\int \frac {\sec^2 t}{|\sec t|} \mathrm{d}t \\
&=\int |\sec t| \mathrm{d}t \\
&=\ln |\sec t + \tan t| + C \end{aligned}$
And reverse the substitutions.
$\begin{aligned} \ln |\sec (\arctan u) + \tan (\arctan u)| + C
&=\ln |\sqrt {1+u^2} + u| + C \\
&=\ln |\sqrt {1+\sin^2 x} + \sin x| + C \end{aligned}$
Or
$u = \sinh t ,\mathrm{d}u = \cosh t$
$\begin{aligned} \implies \int \frac {\cosh t}{\sqrt {1+\sinh^2 t}} \mathrm{d}t
&=\int 1~ \mathrm{d}t \\
&=t + C \\
&=\sinh^{-1} (\sin x) + C \end{aligned}$