The Notes: `int_0^1 x^3 sqrt(1 - x^2) dx` Evaluate the integral

Thursday, March 10, 2016

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

You need to use the following substitution, such that:

$\displaystyle{x}={\sin{{t}}}\Rightarrow{\left.{d}{x}\right.}={\cos{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{1}-{{x}}^{{2}}={1}-{{\sin}}^{{2}}{t}={{\cos}}^{{2}}{t}$

Changing the variable, yields:

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}={\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{{\sin}}^{{3}}{t}\cdot\sqrt{{{{\cos}}^{{2}}{t}}}\cdot{\cos{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{{\sin}}^{{3}}{t}\cdot\sqrt{{{{\cos}}^{{2}}{t}}}\cdot{\cos{{t}}}{\left.{d}{t}\right.}={\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{{\sin}}^{{3}}{t}\cdot{\cos{{t}}}\cdot{\cos{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{{\sin}}^{{3}}{t}\cdot{{\cos}}^{{2}}{t}{\left.{d}{t}\right.}={\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{{\sin}}^{{2}}{t}\cdot{{\cos}}^{{2}}{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{{\sin}}^{{2}}{t}\cdot{{\cos}}^{{2}}{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}={\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}{\left({1}-{{\cos}}^{{2}}{t}\right)}\cdot{{\cos}}^{{2}}{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}$

You need to make the next substitution, such that:

$\displaystyle{\cos{{t}}}={u}\Rightarrow-{\sin{{t}}}{\left.{d}{t}\right.}={d}{u}$

$\displaystyle{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}{\left({1}-{{u}}^{{2}}\right)}\cdot{{u}}^{{2}}\cdot{\left(-{d}{u}\right)}=-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}{{u}}^{{2}}{d}{u}+{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}{{u}}^{{4}}{d}{u}$

$\displaystyle{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}{\left({1}-{{u}}^{{2}}\right)}\cdot{{u}}^{{2}}\cdot{\left(-{d}{u}\right)}={\left(-\frac{{{\left({\cos{{t}}}\right)}}^{{3}}}{{3}}+\frac{{{\left({\cos{{t}}}\right)}}^{{5}}}{{5}}\right)}{{\mid}_{{{t}_{{1}}}}^{{{t}_{{2}}}}}$

Since $\displaystyle{x}={\sin{{t}}}\Rightarrow{t}={\arcsin{{x}}}$

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}={\left(-\frac{{{\left({\cos{{\left({\arcsin{{x}}}\right)}}}\right)}}^{{3}}}{{3}}+{\left(\frac{{{\cos{{\left({\arcsin{{x}}}\right)}}}}^{{5}}}{{5}}\right)}{{\mid}_{{{0}}}^{{{1}}}}\right.}$

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}={\left(-\frac{{{\left({\cos{{\left({\arcsin{{1}}}\right)}}}\right)}}^{{3}}}{{3}}+{\left(\frac{{{\cos{{\left({\arcsin{{1}}}\right)}}}}^{{5}}}{{5}}+\frac{{{\left({\cos{{\left({\arcsin{{0}}}\right)}}}\right)}}^{{3}}}{{3}}-\frac{{{\left({\cos{{\left({\arcsin{{0}}}\right)}}}\right)}}^{{5}}}{{5}}\right)}\right.}$

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=-\frac{{{\left({\cos{{\left(\frac{\pi}{{2}}\right)}}}\right)}}^{{3}}}{{3}}+\frac{{{\left({\cos{{\left(\frac{\pi}{{2}}\right)}}}\right)}}^{{5}}}{{5}}+\frac{{{\left({\cos{{0}}}\right)}}^{{3}}}{{3}}-\frac{{{\left({\cos{{0}}}\right)}}^{{5}}}{{5}}$

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{3}}-\frac{{1}}{{5}}$

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{2}}{{15}}$

Hence, evaluating the definite integral yields $\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\cdot\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{2}}{{15}}.$

The Notes

Thursday, March 10, 2016

$\displaystyle{\int_{{0}}^{{1}}}{{x}}^{{3}}\sqrt{{{1}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

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