Trig Substitutions — Question 8

Question 8 – Solution

Let $$x=3\sec\theta,\qquad dx=3\sec\theta\tan\theta\,d\theta.$$ For $x\ge3$, take $0\le\theta<\pi/2$. Then $$\sqrt{x^2-9}=3\tan\theta.$$ Transform the bounds: $$x=3\Rightarrow\sec\theta=1\Rightarrow\theta=0, \qquad x=5\Rightarrow\cos\theta=\frac35\Rightarrow \theta=\arccos\left(\frac35\right).$$ Thus, $$\begin{align*} I&=\int_0^{\arccos(3/5)} \frac{3\tan\theta}{3\sec\theta} (3\sec\theta\tan\theta)\,d\theta\\ &=3\int_0^{\arccos(3/5)}\tan^2\theta\,d\theta\\ &=3\left[\tan\theta-\theta\right]_0^{\arccos(3/5)}. \end{align*}$$ At the upper bound, a $3$-$4$-$5$ triangle gives $\tan\theta=4/3$. Therefore, $$\begin{align*} I&=3\left(\frac43-\arccos\left(\frac35\right)\right)\\ &=4-3\arccos\left(\frac35\right). \end{align*}$$ Hence, $$\boxed{\displaystyle I=4-3\arccos\left(\frac35\right)}.$$