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,

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

We now use another trigonometric identity,

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

to obtain

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

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,

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