Trig Substitutions — Question 5

Question 5 – Solution

Complete the square inside the radical: $$\begin{align*} 8x-x^2&=-(x^2-8x)\\ &=-\bigl((x-4)^2-16\bigr)\\ &=16-(x-4)^2. \end{align*}$$ Let $$x-4=4\sin\theta,\qquad dx=4\cos\theta\,d\theta.$$ Choose $-\pi/2<\theta<\pi/2$, so $\cos\theta>0$. Then $$\begin{align*} \sqrt{16-(x-4)^2} &=\sqrt{16-16\sin^2\theta}\\ &=4\cos\theta. \end{align*}$$ Thus, $$\begin{align*} I&=\int\frac{4\cos\theta\,d\theta}{4\cos\theta}\\ &=\int d\theta=\theta+C. \end{align*}$$ Since $\sin\theta=(x-4)/4$, $$\boxed{\displaystyle I=\arcsin\left(\frac{x-4}{4}\right)+C}.$$ This formula applies on intervals inside $0<x<8$, where the radical is real and nonzero.