Question

Show that $\int \sqrt{4+8x-5x^2}dx=\sqrt{5}[\frac{5x-4}{10\sqrt{\left(5\right)}}\sqrt{4+8x-5x^2}+\frac{18}{25}{\sin }^{-1}\frac{5x-4}{6}].$

Answer 1

Let $I=\int \sqrt{4+8x-5x^2}dx$

$=\sqrt{5}\sqrt{[-(x^2-\frac{8}{5}x-\frac{4}{5})]}dx$

$=\sqrt{5}\int \sqrt{[-(x^2-\frac{8}{5}x+\frac{16}{25}-\frac{4}{5}-\frac{16}{25})]}dx$

$=\sqrt{5}\int \sqrt{[{\left(\frac{6}{5}\right)}^2-{(x-\frac{4}{5})}^2]}dx$

$=\sqrt{5}[\frac{x-\frac{4}{5}}{2}\sqrt{\{{\left(\frac{6}{5}\right)}^2-{(x-\frac{4}{5})}^2\}}+\frac{1}{2}{\left(\frac{6}{5}\right)}^2{\sin }^1\left(\frac{x-\frac{4}{5}}{\frac{6}{5}}\right)]+C$

$=\sqrt{5}[\frac{5x-4}{10\sqrt{\left(5\right)}}\sqrt{4+8x-5x^2}+\frac{18}{25}{\sin }^{-1}\left(\frac{5x-4}{6}\right)]+C$