Question

${\int }_4^9\frac{dx}{\sqrt{(9-x)(x-4)}}$.

Answer 1

$I={\int }_4^9\frac{dx}{\sqrt{(9-x)(x-4)}}=$ ${\int }_4^9\frac{dx}{\sqrt{(9-x)(x-4)}}$

$={\int }_4^9\frac{dx}{\sqrt{-x^2+13x-36}}$

$={\int }_4^9\frac{dx}{\sqrt{-(x^2-2\ast \frac{13}{2}x+\frac{169}{4})+\frac{169}{4}-36}}$

$={\int }_4^9\frac{dx}{\sqrt{\frac{25}{4}-(x-\frac{13}{2}{)}^2}}$

$={\int }_4^9\frac{dx}{\sqrt{{\left(\frac{5}{2}\right)}^2-{(x-\frac{13}{2})}^2}}$

We know,

$\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+c$

$⟹$ $I={\sin }^{-1}\frac{9-\frac{13}{2}}{5/2}$ $-{\sin }^{-1}\frac{4-\frac{13}{2}}{5/2}$

$⟹$ $I={\sin }^{-1}1-{\sin }^{-1}-1$

$⟹$ $I=\pi /2-(-\pi /2)$

$⟹$ $I=\pi$