Question

Integral of 1/sqrt (81-225x^2)

Answer 1

$\displaystyle\int\frac{\mathrm dx}{\sqrt{81-225x^2}}$

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

$\dfrac{\arcsin\frac{15x}9}{15}+C=\dfrac{\arcsin\frac{5x}3}{15}+C$