Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 03/Solution 1

If we imagine that the integrand is the side of a right-angled triangle, then the other sides of the triangle have length $x$ and 3, and 3 is the length of the hypotenuse. We use this picture to guess the following substitution:

$$\begin{array}{rl}\frac{x}{3}& =\sin (\theta )\\ \mathrm{d}x& =3\cos (\theta )\mathrm{d}\theta \end{array}$$

We thus have

$$\int \sqrt{9-x^2}\mathrm{d}x=\int \sqrt{9-9{\sin }^2(\theta )}\cdot 3\cos (\theta )\mathrm{d}\theta =\int 9{\cos }^{}(\theta ).\mathrm{d}\theta$$

We now use the trigonometric identity ${\cos }^2(\theta )=\frac{1+\cos (2\theta )}{2}$ to evaluate

$$\int 9{\cos }^2(\theta )\mathrm{d}\theta =\frac{9}{2}\int (1+\cos (2\theta ))\mathrm{d}\theta =\frac{9}{2}(\theta +\frac{\sin (2\theta )}{2})+C.$$

We have $\theta =\arcsin \left(\frac{x}{3}\right)$, so, we can rewrite the answer in terms of $x$:

$$\int \sqrt{9-x^2}\mathrm{d}x=\frac{9}{2}(\arcsin \left(\frac{x}{3}\right)+\frac{\sin (2\arcsin \left(\frac{x}{3}\right))}{2})+C=\frac{9}{2}\arcsin \left(\frac{x}{3}\right)+\frac{9}{4}(2\sin (\arcsin \left(\frac{x}{3}\right))\cos (\arcsin \left(\frac{x}{3}\right)))+C,$$

where, in the last step, we used the trigonometric formula $\sin (\alpha +\beta )=\sin (\alpha )\cos (\beta )+\sin (\beta )\cos (\alpha )$ with $\alpha =\beta =\arcsin (x/3)$. By definition of the arcsine, $\sin (\arcsin \left(\frac{x}{3}\right))=\frac{x}{3}$. But this is also to be expected, given our interpretation of $x$ as the length of the side opposite to $\theta$ in a right-angled triangle with hypotenuse of length 3. This picture also shows that

$$\cos (\arcsin \left(\frac{x}{3}\right))=\frac{\sqrt{9-x^2}}{3}.$$

Putting it all together, we find

$$\int \sqrt{9-x^2}\mathrm{d}x=\frac{9}{2}\arcsin \left(\frac{x}{3}\right)+\frac{9}{2}(\frac{x}{3}\cdot \frac{\sqrt{9-x^2}}{3})+C,$$

which is an equation that makes sense for $-3\le x\le 3$.