Trig Substitutions — Question 4

Question 4 – Solution

Because the radical has the form $\sqrt{x^2-16}$, let $$x=4\sec\theta,\qquad dx=4\sec\theta\tan\theta\,d\theta.$$ Since $x>4$, take $0<\theta<\pi/2$. Then $$\sqrt{x^2-16}=\sqrt{16\sec^2\theta-16}=4\tan\theta.$$ Therefore, $$\begin{align*} I&=\int\frac{4\sec\theta\tan\theta\,d\theta} {(16\sec^2\theta)(4\tan\theta)}\\ &=\frac1{16}\int\cos\theta\,d\theta\\ &=\frac1{16}\sin\theta+C. \end{align*}$$ From the reference triangle, $$\sec\theta=\frac{x}{4},\qquad \sin\theta=\frac{\sqrt{x^2-16}}{x}.$$ Hence, $$\boxed{\displaystyle I=\frac{\sqrt{x^2-16}}{16x}+C}.$$ Differentiation recovers $1/(x^2\sqrt{x^2-16})$.