Question

Integrate the function: $\frac {1}{\sqrt {7-6x-x^2}}$

Answer 1

7-6x-x2 can be written as 7-(x2+6x+9-9).
Therefore,
7-(x2+6x+9-9)
= 16-(x2+6x+9)
= 16-(x+3)2
= (4)2-(x+3)2
∴ ∫$$\frac {1}{\sqrt {7-6x-x^2}}\ dx = $∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx$
Let x+3 = t
⇒ dx = dt
⇒ $∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx$ = $∫\frac {1}{\sqrt {(4)^2-t^2}} dt$
= $sin^{-1}(\frac t4)+C$
=$sin^{-1}(\frac {x+3}4)+C$