Integrals Involving Quadratics — Question 8

Question 8 – Solution

Step 1: Complete the square. $$\begin{align*} 6x-x^2 &=-(x^2-6x)\\ &=-\bigl((x-3)^2-9\bigr)\\ &=9-(x-3)^2. \end{align*}$$ Step 2: Determine the domain. The radicand must be nonnegative: $$\begin{align*} 9-(x-3)^2&\ge0,\\ (x-3)^2&\le9,\\ -3\le x-3&\le3,\\ 0\le x&\le6. \end{align*}$$ The denominator cannot equal zero, so the integrand’s domain is $$\boxed{0<x<6}.$$ Step 3: Substitute $u=x-3$, so $du=dx$: $$I=\int\frac{du}{\sqrt{3^2-u^2}}.$$ Step 4: Use the inverse-sine formula. $$\begin{align*} I&=\arcsin\left(\frac{u}{3}\right)+C\\ &=\arcsin\left(\frac{x-3}{3}\right)+C. \end{align*}$$ $$\boxed{\arcsin\frac{x-3}{3}+C}$$