Question

Find the integral
$\int \frac{dx}{\sqrt{9x-4x^2}}$

Answer 1

Consider the given integral.

$I=\int \frac{dx}{\sqrt{9x-4x^2}}$

$I=\int \frac{dx}{\sqrt{9x-4x^2+\frac{81}{16}-\frac{81}{16}}}$

$I=\int \frac{dx}{\sqrt{{\left(\frac{9}{4}\right)}^2-{(2x-\frac{9}{4})}^2}}$

Put

$t=2x-\frac{9}{4}$

$\frac{dt}{2}=dx$

Therefore,

$I=\frac{1}{2}\int \frac{dt}{\sqrt{{\left(\frac{9}{4}\right)}^2-t^2}}$

$I=\frac{1}{2}{\sin }^{-1}\left(\frac{t}{\frac{9}{4}}\right)+C$

Put the value of t in above expression, we get

$I=\frac{1}{2}{\sin }^{-1}\left(\frac{4(2x-\frac{9}{4})}{9}\right)+C$

$I=\frac{1}{2}{\sin }^{-1}\left(\frac{(8x-9)}{9}\right)+C$

Hence, the value of this integral is $\frac{1}{2}{\sin }^{-1}\left(\frac{(8x-9)}{9}\right)+C$.