Question

Integrate the function: \(\frac {1}{\sqrt {8+3x-x^2}}\)

Answer 1

8+3x-x² can be written as 8 - (x² - 3x + \(\frac 94\) - \(\frac 94\)).

Therefore, 

8 - (x²- 3x + \(\frac 94\) - \(\frac 94\))

= \(\frac {41}{4}\) - (x-\(\frac 32\))²

⇒ \(∫\)\(\frac {1}{\sqrt {8+3x-x^2}}\ dx\)  = \(∫\frac {1}{\sqrt {\frac {41}{4}-(x-\frac 32)^2}} dx\)

Let x-\(\frac 32\) = t

∴ dx = dt

⇒ \(∫\frac {1}{\sqrt {\frac {41}{4}-(x-\frac 32)^2}}\  dx\) = \(∫\frac {1}{\sqrt {(\frac {\sqrt {41}}{2})^2-t^2}} \ dt\)

= \(sin^{-1}(\frac {t}{\frac {\sqrt {41}}{2}})+ C\)

= \(sin^{-1}(\frac {x-\frac 32}{\frac {\sqrt {41}}{2}})+ C\)

= \(sin^{-1}(\frac {2x-3}{\sqrt {41}})+C\)