\(\displaystyle\int\frac{\mathrm dx}{\sqrt{81-225x^2}}\)
Let \(x=\dfrac9{15}\sin t\), so that \(\mathrm dx=\dfrac9{15}\cos t\,\mathrm dt\) and the integral becomes
\(\displaystyle\int\frac{\dfrac9{15}\cos t}{\sqrt{81-225\left(\frac9{15}\sin t\right)^2}}\,\mathrm dt\)
\(\displaystyle\int\frac{\dfrac9{15}\cos t}{\sqrt{81-225\times\frac{81}{225}\sin^2 t}}\,\mathrm dt\)
\(\displaystyle\frac1{15}\int\frac{\cos t}{\sqrt{1-\sin^2 t}}\,\mathrm dt\)
\(\displaystyle\frac1{15}\int\frac{\cos t}{\sqrt{\cos^2 t}}\,\mathrm dt\)
\(\displaystyle\frac1{15}\int\frac{\cos t}{|\cos t|}\,\mathrm dt\)
When \(\cos t>0\), you have \(|\cos t|=\cos t\). Under these conditions, you can write
\(\displaystyle\frac1{15}\int\frac{\cos t}{\cos t}\,\mathrm dt=\frac1{15}\int\mathrm dt=\frac t{15}+C\)
and back-substituting yields
\(\dfrac{\arcsin\frac{15x}9}{15}+C=\dfrac{\arcsin\frac{5x}3}{15}+C\)
