Trig Substitutions — Question 9

Question 9 – Solution

Let $$x=a\tan\theta,\qquad dx=a\sec^2\theta\,d\theta.$$ Since $a>0$, choose $-\pi/2<\theta<\pi/2$. Then $$\begin{align*} (a^2+x^2)^{3/2} &=\bigl(a^2+a^2\tan^2\theta\bigr)^{3/2}\\ &=(a^2\sec^2\theta)^{3/2}=a^3\sec^3\theta. \end{align*}$$ Therefore, $$\begin{align*} I&=\int\frac{a\sec^2\theta}{a^3\sec^3\theta}\,d\theta\\ &=\frac1{a^2}\int\cos\theta\,d\theta\\ &=\frac1{a^2}\sin\theta+C. \end{align*}$$ From a reference triangle, $$\tan\theta=\frac{x}{a},\qquad \sin\theta=\frac{x}{\sqrt{a^2+x^2}}.$$ Thus, $$\boxed{\displaystyle I=\frac{x}{a^2\sqrt{a^2+x^2}}+C}.$$ Indeed, $$\frac{d}{dx}\left(\frac{x}{a^2\sqrt{a^2+x^2}}\right) =\frac1{(a^2+x^2)^{3/2}},$$ which verifies the result.