Science:Math Exam Resources/Courses/MATH105/April 2016/Question 01 (k)/Solution 1

Let $x=4\sin(u)$. We then use the double-angle formulas and the fact that $dx=4\cos(u)du$ to get

${\begin{aligned}\int {\sqrt {16-x^{2}}}dx&=\int {\sqrt {16(1-\sin ^{2}(u))}}\ 4\cos(u)du=\\&=\int 16{\sqrt {\cos ^{2}(u)}}\cos(u)du\\&=\int 16\cos ^{2}(u)du\\&=16\int {\frac {1+\cos(2u)}{2}}du\\&=16({\frac {u}{2}}+{\frac {\sin(2u)}{4}})+C\\&=8u+4\sin(2u)+C\\&=8u+8\sin(u)\cos(u)+C.\end{aligned}}$

Now, we should find $u$, $\sin u$ and $\cos u$ in terms of $x$:

We know that ${\frac {x}{4}}=\sin(u)$, so $u=\sin ^{-1}{\bigg (}{\frac {x}{4}}{\bigg )}=\arcsin {\bigg (}{\frac {x}{4}}{\bigg )}.$ Also, we get $\cos(u)={\sqrt {1-\sin ^{2}u}}={\sqrt {1-{\frac {x^{2}}{16}}}}$. Thus,

$8u=8\sin ^{-1}{\bigg (}{\frac {x}{4}}{\bigg )}$ and $8\sin(u)\cos(u)=8\times {\frac {x}{4}}\times {\sqrt {1-{\frac {x^{2}}{16}}}}$,

Therefore, the final answer is:

$\int {\sqrt {16-x^{2}}}dx={\color {Blue}8\sin ^{-1}{\bigg (}{\frac {x}{4}}{\bigg )}+2x{\sqrt {1-{\frac {x^{2}}{16}}}}}+C.$