Trig Substitutions — Question 3

Question 3

Find the total area under $$y=(x^2+9)^{-3/2}$$ on $0\le x<\infty$.

Question 3 – Solution

The required area is the improper integral $$I=\lim_{b\to\infty}\int_0^b\frac{dx}{(x^2+9)^{3/2}}.$$ Let $x=3\tan\theta$. Then $$dx=3\sec^2\theta\,d\theta,\qquad (x^2+9)^{3/2}=27\sec^3\theta.$$ The bounds transform as $x=0\Rightarrow\theta=0$ and $x\to\infty\Rightarrow\theta\to\pi/2$. Thus, $$\begin{align*} I&=\frac19\int_0^{\pi/2}\cos\theta\,d\theta\\ &=\frac19\left[\sin\theta\right]_0^{\pi/2}\\ &=\frac19. \end{align*}$$ The integrand is positive, so this convergent improper integral equals the geometric area. Therefore, $$\boxed{I=\frac19}.$$