Question

Integrate the following w.r.t $x$
$\frac{x^2-3x+1}{\sqrt{1-x^2}}$

Answer 1

$I=\int \frac{x^2-3x+1}{\sqrt{1-x_2}}dx$

put $x=\sin \theta$

$dx=\cos \theta d\theta$

$=\int \frac{({\sin }^2\theta -3\sin \theta +1)}{\cos \theta }\times \cos \theta d\theta$

$=\int (\frac{1-\cos 2\theta }{2}+1-3\sin \theta )d\theta$

$=\frac{\theta }{2}\frac{-1}{2}\frac{\sin 2\theta }{2}+\theta +3\cos \theta +C$

$=\frac{-1}{4}\sin 2\theta +3\cos \theta +\frac{3\theta }{2}+C$

$=\frac{-1}{4}\times 2\times x\sqrt{1-x^2}+3\sqrt{1-x^2}+\frac{3}{2}{\sin }^{-1}x+C$

$=\frac{-x}{2}\sqrt{1-x^2}+3\sqrt{1-x^2}+\frac{3}{2}{\sin }^{-1}x+C$