The Notes: `int_0^a x^2 sqrt(a^2 - x^2) dx` Evaluate the integral

Thursday, March 26, 2009

$\displaystyle{\int_{{0}}^{{a}}}{{x}}^{{2}}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{a}}}{{x}}^{{2}}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

Let's evaluate the indefinite integral by applying integral substitution,

Let $\displaystyle{x}={a}{\sin{{\left({u}\right)}}}$

$\displaystyle{\left.{d}{x}\right.}={a}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle\int{{x}}^{{2}}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\int{{\left({a}{\sin{{\left({u}\right)}}}\right)}}^{{2}}\sqrt{{{{a}}^{{2}}-{{\left({a}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}}{a}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\int{{a}}^{{2}}{{\sin}}^{{2}}{\left({u}\right)}\sqrt{{{{a}}^{{2}}-{{a}}^{{2}}{{\sin}}^{{2}}{\left({u}\right)}}}{a}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle={{a}}^{{3}}\int{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}\sqrt{{{{a}}^{{2}}{\left({1}-{{\sin}}^{{2}}{\left({u}\right)}\right)}}}{d}{u}$

$\displaystyle={{a}}^{{3}}\int{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}{a}\sqrt{{{1}-{{\sin}}^{{2}}{\left({u}\right)}}}{d}{u}$

Now use the identity: $\displaystyle{1}-{{\sin}}^{{2}}{\left({x}\right)}={{\cos}}^{{2}}{\left({x}\right)}$

$\displaystyle={{a}}^{{4}}\int{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}\sqrt{{{{\cos}}^{{2}}{\left({u}\right)}}}{d}{u}$

$\displaystyle={{a}}^{{4}}\int{{\sin}}^{{2}}{\left({u}\right)}{{\cos}}^{{2}}{\left({u}\right)}{d}{u}$

Now use the identity: $\displaystyle{{\cos}}^{{2}}{\left({x}\right)}{{\sin}}^{{2}}{\left({x}\right)}=\frac{{{1}-{\cos{{\left({4}{x}\right)}}}}}{{8}}$

$\displaystyle={{a}}^{{4}}\int\frac{{{1}-{\cos{{\left({4}{u}\right)}}}}}{{8}}{d}{u}$

$\displaystyle=\frac{{{a}}^{{4}}}{{8}}\int{\left({1}-{\cos{{\left({4}{u}\right)}}}{d}{u}\right.}$

$\displaystyle=\frac{{{a}}^{{4}}}{{8}}{\left(\int{1}{d}{u}-\int{\cos{{\left({4}{u}\right)}}}{d}{u}\right)}$

$\displaystyle=\frac{{{a}}^{{4}}}{{8}}{\left({u}-\frac{{\sin{{\left({4}{u}\right)}}}}{{4}}\right)}$

Substitute back $\displaystyle{u}={\arcsin{{\left(\frac{{x}}{{a}}\right)}}}$

$\displaystyle=\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left(\frac{{x}}{{a}}\right)}}}-\frac{{\sin{{\left({4}{\arcsin{{\left(\frac{{x}}{{a}}\right)}}}\right)}}}}{{4}}\right)}$

add a constant C to the solution,

$\displaystyle=\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left(\frac{{x}}{{a}}\right)}}}-\frac{{1}}{{4}}{\sin{{\left({4}{\arcsin{{\left(\frac{{x}}{{a}}\right)}}}\right)}}}\right)}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{a}}}{{x}}^{{2}}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}={{\left[\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left(\frac{{x}}{{a}}\right)}}}-\frac{{1}}{{4}}{\sin{{\left({4}{\arcsin{{\left(\frac{{x}}{{a}}\right)}}}\right)}}}\right)}\right]}_{{0}}^{{a}}}$

$\displaystyle={\left[\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left(\frac{{a}}{{a}}\right)}}}-\frac{{1}}{{4}}{\sin{{\left({4}{\arcsin{{\left(\frac{{a}}{{a}}\right)}}}\right)}}}\right)}\right]}-{\left[\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left(\frac{{0}}{{a}}\right)}}}-\frac{{1}}{{4}}{\sin{{\left({4}{\arcsin{{\left(\frac{{0}}{{a}}\right)}}}\right)}}}\right)}\right]}$

$\displaystyle={\left[\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left({1}\right)}}}-\frac{{1}}{{4}}{\sin{{\left({4}{\arcsin{{\left({1}\right)}}}\right)}}}\right)}\right]}-{\left[\frac{{{a}}^{{4}}}{{8}}{\left({\arcsin{{\left({0}\right)}}}-\frac{{1}}{{4}}{\sin{{\left({4}{\arcsin{{\left({0}\right)}}}\right)}}}\right)}\right]}$

$\displaystyle={\left[\frac{{{a}}^{{4}}}{{8}}{\left(\frac{\pi}{{2}}-\frac{{1}}{{4}}{\sin{{\left({4}\cdot\frac{\pi}{{2}}\right)}}}\right)}\right]}-{\left[{a}\frac{{4}}{{8}}{\left({0}-\frac{{1}}{{4}}{\sin{{\left({4}\cdot{0}\right)}}}\right)}\right]}$