{\int_{{0}}^{{a}}}{3}{x}^{2}\sqrt{{{a}^{2}-{x}^{2}}}{\left.{d}{x}\right.}

mattgondek4 2021-09-02 Answered

\({\int_{{0}}^{{a}}}{3}{x}^{2}\sqrt{{{a}^{2}-{x}^{2}}}{\left.{d}{x}\right.}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Bertha Stark
Answered 2021-09-03 Author has 3769 answers

Take the integral:
\(\int{3}{x}^{2}\sqrt{{{a}^{2}-{x}^{2}}}{\left.{d}{x}\right.}\)
Factor out constants:
\(\int{3}{x}^{2}\sqrt{{{a}^{2}-{x}^{2}}}{\left.{d}{x}\right.}\)
For the integrand \({x}^{2}\sqrt{{{a}^{2}-{x}^{2}}}\), (assuming all variables are positive) substitute \({x}={a} \sin{{\left({u}\right)}}{\quad\text{and}\quad}{\left.{d}{x}\right.}={a} \cos{{\left({u}\right)}}{d}{u}\). Then \(\sqrt{{{a}^{2}-{x}^{2}}}=\sqrt{{{a}^{2}-{a}^{2}{{\sin}^{2}{\left({u}\right)}}}}={a} \cos{{\left({u}\right)}}{\quad\text{and}\quad}{u}={{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}\):
\(={3}{a}\int{a}^{3}{{\sin}^{2}{\left({u}\right)}}{{\cos}^{2}{\left({u}\right)}}{d}{u}\)
Factor out constants:
\(={3}{a}^{4}\int{{\sin}^{2}{\left({u}\right)}}{{\cos}^{2}{\left({u}\right)}}{d}{u}\)
Write \({{\cos}^{2}{\left({u}\right)}}{a}{s}{1}-{{\sin}^{2}{\left({u}\right)}}\):
\(={3}{a}^{4}\int{{\sin}^{2}{\left({u}\right)}}{\left({1}-{{\sin}^{2}{\left({u}\right)}}\right)}{d}{u}\)
Expanding the integrand \({{\sin}^{2}{\left({u}\right)}}{\left({1}-{{\sin}^{2}{\left({u}\right)}}\right)}\) gives \({{\sin}^{2}{\left({u}\right)}}-{{\sin}^{4}{\left({u}\right)}}\):
\(={3}{a}^{4}\int{\left({{\sin}^{2}{\left({u}\right)}}-{{\sin}^{4}{\left({u}\right)}}\right)}{d}{u}\)
Integrate the sum term by term and factor out constants:
\(=-{3}{a}^{4}\int{{\sin}^{4}{\left({u}\right)}}{d}{u}+{3}{a}^{4}\int{{\sin}^{2}{\left({u}\right)}}{d}{u}\)
Use the reduction formula, \(\int{{\sin}^{m}{\left({u}\right)}}{d}{u}=-\frac{{ \cos{{\left({u}\right)}}{{\sin}^{{{m}-{1}}}{\left({u}\right)}}}}{{m}}+\frac{{{m}-{1}}}{{m}}\int{{\sin}^{{-{2}+{m}}}{\left({u}\right)}}{d}{u}\), where m = 4:
\(=\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}+\frac{{{3}{a}^{4}}}{{4}}\int{{\sin}^{2}{\left({u}\right)}}{d}{u}\)
Write \({{\sin}^{2}{\left({u}\right)}}{a}{s}\frac{1}{{2}}-\frac{1}{{2}} \cos{{\left({2}{u}\right)}}\):
\(=\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}+\frac{{{3}{a}^{4}}}{{4}}\int{\left(\frac{1}{{2}}-\frac{1}{{2}} \cos{{\left({2}{u}\right)}}\right)}{d}{u}\)
Integrate the sum term by term and factor out constants:
\(=\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}-\frac{{{3}{a}^{4}}}{{8}}\int \cos{{\left({2}{u}\right)}}{d}{u}+\frac{{{3}{a}^{4}}}{{8}}\int{1}{d}{u}\)
For the integrand cos(2 u), substitute \(s = 2 u\) and \(ds = 2 du\):
\(=\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}-\frac{{{3}{a}^{4}}}{{16}}\int \cos{{\left({s}\right)}}{d}{s}+\frac{{{3}{a}^{4}}}{{8}}\int{1}{d}{u}\)
The integral of \(\cos(s)\) is \(\sin(s)\):
\(=-\frac{3}{{16}}{a}^{4} \sin{{\left({s}\right)}}+\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}+\frac{{{3}{a}^{4}}}{{8}}\int{1}{d}{u}\)
The integral of 1 is u:
\(=-\frac{3}{{16}}{a}^{4} \sin{{\left({s}\right)}}+\frac{{{3}{a}^{4}{u}}}{{8}}+\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}+\text{constant}\)
Substitute back for \(s = 2 u\):
\(=\frac{{{3}{a}^{4}{u}}}{{8}}+\frac{3}{{4}}{a}^{4}{{\sin}^{3}{\left({u}\right)}} \cos{{\left({u}\right)}}-\frac{3}{{8}}{a}^{4} \sin{{\left({u}\right)}} \cos{{\left({u}\right)}}+\text{constant}\)
Substitute back for \({u}={{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}\):
\(=\frac{3}{{8}}{a}^{4}{{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}+\frac{3}{{4}}{a}^{4}{ \sin{{\left({{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}\right)}}^{3} \cos{{\left({{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}\right)}}}-\frac{3}{{8}}{a}^{4} \sin{{\left({{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}\right)}} \cos{{\left({{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}\right)}}+\text{constant}\)
Simplify using \(\cos{{\left({{\sin}^{{-{1}}}{\left({z}\right)}}\right)}}=\sqrt{{{1}-{z}^{2}}}{\quad\text{and}\quad} \sin{{\left({{\sin}^{{-{1}}}{\left({z}\right)}}\right)}}={z}\):
\(=\frac{3}{{8}}{a}^{4}{{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}-\frac{3}{{8}}{a}^{2}{x}\sqrt{{{a}^{2}-{x}^{2}}}+\frac{3}{{4}}{x}^{3}\sqrt{{{a}^{2}-{x}^{2}}}+\text{constant}\)
Factor the answer a different way:
\(=\frac{1}{{8}}{\left({3}{a}^{4}{{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}+{3}{x}\sqrt{{{a}^{2}-{x}^{2}}}{\left({2}{x}^{2}-{a}^{2}\right)}\right)}+\text{constant}\)
Which is equivalent for restricted x and a values to:
Answer:
\(=\frac{3}{{8}}\sqrt{{{a}^{2}-{x}^{2}}}{\left(-{a}^{2}{x}+\frac{{{a}^{3}{{\sin}^{{-{1}}}{\left(\frac{x}{{a}}\right)}}}}{\sqrt{{{1}-\frac{{x}^{2}}{{a}^{2}}}}}+{2}{x}^{3}\right)}+\text{constant}\)

Have a similar question?
Ask An Expert
49
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...