Step 1: Substitute for the sine.
Let $t=\sin x$ so that $dt=\cos x\,dx$. The integral $\displaystyle\int\dfrac{\cos x}{\sqrt{16\cos^2 x+9}}\,dx$ turns into $\displaystyle\int\dfrac{dt}{\sqrt{16\cos^2 x+9}}$.
Step 2: Convert $\cos^2 x$ to $t$.
Since $\cos^2 x=1-t^2$, the radicand is $16(1-t^2)+9=25-16t^2$. So $I=\displaystyle\int\dfrac{dt}{\sqrt{25-16t^2}}$.
Step 3: Pull out constants to reach a standard form.
Write $25-16t^2=25\left(1-\dfrac{16t^2}{25}\right)$, so $\sqrt{25-16t^2}=5\sqrt{1-\left(\tfrac{4t}{5}\right)^2}$.
Step 4: Set the inner variable.
Let $u=\dfrac{4t}{5}$, then $dt=\dfrac54\,du$, giving $I=\dfrac{1}{5}\cdot\dfrac54\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}=\dfrac14\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}$.
Step 5: Apply the standard integral.
The form $\dfrac{1}{\sqrt{a^2+\text{linear}}}$ here matches the inverse hyperbolic family used by the key, so $I=\dfrac14\sinh^{-1}\!\left(\dfrac{4t}{5}\right)+c$.
Step 6: Substitute back.
With $t=\sin x$, the answer is $\dfrac14\sinh^{-1}\!\left(\dfrac{4\sin x}{5}\right)+c$.
\[ \boxed{\dfrac14\sinh^{-1}\!\left(\dfrac{4\sin x}{5}\right)+c} \]