Question

Evaluate: $\int (3-2x).\sqrt{2+x-x^2}dx.$

Answer 1

$\int (3-2x)\sqrt{2+x-x^2}dx=\int (3-2x)\sqrt{{\left(\frac{3}{2}\right)}^2-{(x-\frac{1}{2})}^2}dx=\frac{3}{2}\int (3-2x)\sqrt{1-{\left(\frac{2x-1}{3}\right)}^2}dx$

Substituting $\frac{2x-1}{3}=\cos \theta \Rightarrow dx=-\frac{3}{2}\sin \theta d\theta$ , we get:

$\frac{3}{2}\int (3-2x)\sqrt{1-{\left(\frac{2x-1}{3}\right)}^2}dx=\frac{3}{2}\int (2-3\cos \theta )\sin \theta (-\frac{3}{2}\sin \theta )d\theta =\frac{9}{4}\int (3\cos \theta -2){\sin }^2\theta d\theta =\frac{27}{4}\int \cos \theta {\sin }^2\theta d\theta -\frac{9}{2}\int {\sin }^2\theta d\theta =\frac{9}{4}{\sin }^3\theta -\frac{9}{4}\int (1-\cos 2\theta )d\theta =\frac{9}{4}{\sin }^3\theta -\frac{9}{4}\theta +\frac{9}{8}\sin 2\theta +C$

Substituting back, we get:

$\frac{9}{4}{\sin }^3\theta -\frac{9}{4}\theta +\frac{9}{8}\sin 2\theta +C=\frac{9}{4}{(1-{\left(\frac{2x-1}{3}\right)}^2)}^{\frac{3}{2}}-\frac{9}{4}{\cos }^{-1}\frac{2x-1}{3}+\frac{9}{4}\left(\frac{2x-1}{3}\right){(1-{\left(\frac{2x-1}{3}\right)}^2)}^{\frac{1}{2}}+C$