Science:Math Exam Resources/Courses/MATH101 C/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:

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

We thus have

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

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

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

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

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

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, $\sin \left(\arcsin \left({\frac {x}{3}}\right)\right)={\frac {x}{3}}$. 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

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

Putting it all together, we find

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

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