Question

Evaluate :
${\int }_0^4{(16-x^2)}^{1/2}dx$

Answer 1

${\int }_0^4{(16-x^2)}^{\frac{1}{2}}dx$

Take $x=4\cos \theta \Rightarrow dt=-4\sin \theta d\theta$

When $x=0\Rightarrow \theta =\frac{\pi }{2}$

When $x=4\Rightarrow 4\cos \theta =4\Rightarrow \theta =0$

$={\int }_{\frac{\pi }{2}}^0{(16-{16\cos }^2\theta )}^{\frac{1}{2}}(-4\sin \theta )d\theta$

$=4{\int }_{\frac{\pi }{2}}^0{\left({\sin }^2\theta \right)}^{\frac{1}{2}}(-4\sin \theta )d\theta$

$=-16{\int }_{\frac{\pi }{2}}^0{\sin }^2\theta d\theta$

$=\frac{16}{2}{\int }_0^{\frac{\pi }{2}}2{\sin }^2\theta d\theta$

$=8{\int }_0^{\frac{\pi }{2}}(1-\cos 2\theta )d\theta$

$=8{[\theta -\frac{\sin 2\theta }{2}]}_0^{\frac{\pi }{2}}$

$=8[\frac{\pi }{2}-0-\frac{1}{2}(\sin \pi -\sin 0)]$

$=8[\frac{\pi }{2}-0-\frac{1}{2}(0-0)]$

$=\frac{8\pi }{2}=4\pi$