Science:Math Exam Resources/Courses/MATH101/April 2009/Question 04 (a)/Solution 1

Following the hint, we let

$\displaystyle x=2\sin \theta$

Then

$\displaystyle dx=2\cos \theta d\theta$

and so we have

${\begin{aligned}\int (4-x^{2})^{-3/2}\,dx&=\int {\frac {dx}{{\sqrt {4-x^{2}}}^{3}}}\\&=\int {\frac {2\cos \theta d\theta }{{\sqrt {4-4\sin ^{2}\theta }}^{3}}}\\&=\int {\frac {2\cos \theta d\theta }{{\sqrt {4-4\sin ^{2}\theta }}^{3}}}\\&=\int {\frac {2\cos \theta d\theta }{{\sqrt {4\cos ^{2}\theta }}^{3}}}\\&=\int {\frac {2\cos \theta d\theta }{8\cos ^{3}\theta }}\\&=\int {\frac {d\theta }{4\cos ^{2}\theta }}\\&={\frac {1}{4}}\int \sec ^{2}\theta \\&={\frac {\tan \theta }{4}}+C\end{aligned}}$

To substitute back to an expression with x you need to undo the original substitution $\sin \theta =x/2.$ With the help of a little diagram (included below) and the Pythagorean theorem to see that

$\tan \theta ={\frac {x}{\sqrt {4-x^{2}}}}$

and hence

$\int (4-x^{2})^{-3/2}\,dx={\frac {\tan \theta }{4}}+C={\frac {x}{4{\sqrt {4-x^{2}}}}}+C$

completing the question.