Trig Substitutions — Question 6

Question 6 – Solution

Let $x=2\tan\theta$, with $-\pi/2<\theta<\pi/2$. Then $$dx=2\sec^2\theta\,d\theta,\qquad \sqrt{x^2+4}=2\sec\theta.$$ Therefore, $$I=4\int\sec^3\theta\,d\theta.$$ Using $\int\sec^3\theta\,d\theta =\frac12\bigl(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\bigr)$, $$I=2\sec\theta\tan\theta+2\ln|\sec\theta+\tan\theta|+C.$$ Now $\tan\theta=x/2$ and $\sec\theta=\sqrt{x^2+4}/2$. Hence, $$\begin{align*} I&=\frac{x}{2}\sqrt{x^2+4} +2\ln\left|\frac{x+\sqrt{x^2+4}}2\right|+C\\ &=\frac{x}{2}\sqrt{x^2+4} +2\ln\left|x+\sqrt{x^2+4}\right|+C, \end{align*}$$ where $-2\ln2$ is absorbed into $C$. Thus, $$\boxed{\displaystyle I=\frac{x}{2}\sqrt{x^2+4} +2\ln\left|x+\sqrt{x^2+4}\right|+C}.$$ Differentiating and simplifying gives $I'=\sqrt{x^2+4}$, as required.