Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (d)/Solution 1

Let $x=2\sin \theta$. Then, $dx=2\cos \theta d\theta$. By substitution,

$\begin{array}{rl}\int \sqrt{4-x^2}dx& =\int \sqrt{4-4{\sin }^2\theta }\cdot 2\cos \theta d\theta =\int \sqrt{4{\cos }^2\theta }\cdot 2\cos \theta d\theta \\ & =\int 2\cos \theta \cdot 2\cos \theta d\theta =\int 4{\cos }^{}\theta d\theta .\end{array}$

We now use another trigonometric identity,

$\begin{array}{r}{\cos }^2\theta =\frac{1+\cos 2\theta }{2},\end{array}$

to obtain

$\begin{array}{rl}\int 4{\cos }^2\theta d\theta & =\int 2(1+\cos 2\theta )d\theta \\ & =2\int d\theta +2\int \cos 2\theta d\theta \\ & =2\theta +2\left(\frac{\sin 2\theta }{2}\right)+C\\ & =2\theta +2\sin \theta \cos \theta +C.\end{array}$

In the last step we used that $\sin 2\theta =2\sin \theta \cos \theta$. All that's left to do is to express this result in terms of $x$.

First, notice that $x=2\sin \theta$, $\theta =\arcsin \left(\frac{x}{2}\right)$ and hence $\sin \theta =\frac{x}{2}$. Also, by using the below picture, $\cos \theta =\frac{\sqrt{4-x^2}}{2}$. Thus,

$\begin{array}{r}\int \sqrt{4-x^2}dx=2\theta +2\sin \theta \cos \theta +C=2\arcsin \frac{x}{2}+2\cdot \frac{x}{2}\cdot \frac{\sqrt{4-x^2}}{2}+C=2\arcsin \frac{x}{2}+\frac{x\sqrt{4-x^2}}{2}+C.\end{array}$