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

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

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

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

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