Trig Substitutions — Question 7

Question 7 – Solution

Choose $$\boxed{x=3\tan\theta}.$$ This substitution uses $1+\tan^2\theta=\sec^2\theta$: $$x^2+9=9\tan^2\theta+9=9\sec^2\theta.$$ Taking $-\pi/2<\theta<\pi/2$ gives $\sec\theta>0$, so $$dx=3\sec^2\theta\,d\theta,\qquad \sqrt{x^2+9}=3\sec\theta.$$ Hence, $$\begin{align*} I&=\int\frac{3\sec^2\theta}{3\sec\theta}\,d\theta\\ &=\int\sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|+C. \end{align*}$$ Since $\tan\theta=x/3$ and $\sec\theta=\sqrt{x^2+9}/3$, $$\begin{align*} I&=\ln\left|\frac{x+\sqrt{x^2+9}}3\right|+C\\ &=\ln|x+\sqrt{x^2+9}|+C. \end{align*}$$ The constant $-\ln3$ is absorbed into $C$. Thus, $$\boxed{\displaystyle \int\frac{dx}{\sqrt{x^2+9}} =\ln|x+\sqrt{x^2+9}|+C}.$$ The sine substitution does not turn $x^2+9$ into a single squared trigonometric function, so it does not simplify the radical.