Question

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

Answer 1

We have,

$I=\int (3-2x)\sqrt{2+x-x^2}dx$

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

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

$I=I_1+I_2$

Let,

$I_1=\int (1-2x)\sqrt{2+x-x^2}dx$

$I_2=2\int \sqrt{2+x-x^2}dx$

First,

$I_1=\int (1-2x)\sqrt{2+x-x^2}dx$

Let,

$2+x-x^2=t$

$\frac{d}{dx}(2+x-x^2)=\frac{dt}{dx}$

$(0+1-2x)=\frac{dt}{dx}$

$(1-2x)dx=dt$

$I_1=\int \sqrt{t}dt$

$=\int t^{\frac{1}{2}}dt$

$=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$

$=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C$

$=\frac{2}{3}{t}^{\frac{3}{2}}+C$

$=\frac{2}{3}{(2+x-x^2)}^{\frac{3}{2}}+C$

$I_2=2\sqrt{2+x-x^2}dx$

$=2\sqrt{2-(x^2-x)}dx$

$=2\sqrt{2-(x^2-x+\frac{1}{4}-\frac{1}{4})}dx$

$=2\sqrt{2-{(x-\frac{1}{2})}^2+\frac{1}{4}}dx$

$=2\sqrt{\frac{9}{4}-{(x-\frac{1}{2})}^2}dx$

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

We know that,

$\sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+C$

Then,

$I_2=2\sqrt{{\left(\frac{3}{2}\right)}^2-{(x-\frac{1}{2})}^2}dx$

$I_2=2[\frac{{(x-\frac{1}{2})}^2\sqrt{{\left(\frac{3}{2}\right)}^2-{(x-\frac{1}{2})}^2}}{2}+\frac{{\left(\frac{3}{2}\right)}^2}{2}{\sin }^{-1}\left(\frac{(x-\frac{1}{2})}{\left(dfrac32\right)}\right)]+C$

Now,

$I=I_1+I_2$

$I=\frac{2}{3}{(2+x-x^2)}^{\frac{3}{2}}+2[\frac{{(x-\frac{1}{2})}^2\sqrt{{\left(\frac{3}{2}\right)}^2-{(x-\frac{1}{2})}^2}}{2}+\frac{{\left(\frac{3}{2}\right)}^2}{2}{\sin }^{-1}\left(\frac{(x-\frac{1}{2})}{\left(\frac{3}{2}\right)}\right)]+C$

Hence, this is the answer.