Could someone explain to me how I can find $\int x^3$$\sqrt{1-x^2}\ dx$ ?


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  • $\begingroup$ Do you mean "indefinite" instead of "undefined"? $\endgroup$ – 0XLR Jul 3 at 3:06
  • $\begingroup$ Sorry, I adjusted. $\endgroup$ – Mycroft Jul 3 at 3:07
  • 1
    $\begingroup$ Straightforward integration by parts: $-\frac{1}{15} \left(1-x^2\right)^{3/2} \left(3 x^2+2\right)$. $\endgroup$ – David G. Stork Jul 3 at 3:09
  • $\begingroup$ Could you explain the steps? $\endgroup$ – Mycroft Jul 3 at 3:15

Because the power of $x$ outside the radical is odd, this is most easily accomplished with the substitution $u = 1-x^2$ rather than a trig substitution. Then $du = -2x\,dx$ and \begin{multline} \int x^3\sqrt{1-x^2}dx = -\frac{1}{2}\int(1-u)\sqrt{u}\,du = -\frac{1}{2}\int\left(u^{1/2}-u^{3/2}\right)\,du = \frac{u^{5/2}}{5} - \frac{u^{3/2}}{3}\\ = \frac{(1-x^2)^{5/2}}{5}-\frac{(1-x^2)^{3/2}}{3} = -\frac{2+3x^2}{15}(1-x^2)^{3/2} \end{multline}

  • $\begingroup$ (+1) modulo the constant of integration, this is correct. $\endgroup$ – robjohn Jul 3 at 3:52

Using a u substition is very useful in this case

$$\int x^3\sqrt{1-x^2}dx$$ $$u=1-x^2$$

$$du=-2x\hspace{1mm} dx$$

$$dx=-\frac{1}{2x} du$$

$$\int x^3\sqrt{u} \cdot -\frac{1}{2x}du$$ $$-\frac{1}{2} \int (1-u)\sqrt{u} \hspace{1mm}du$$ $$-\frac{1}{2} \int u^{\frac{1}{2}}-u^{\frac{3}{2}} \hspace{1mm}du$$ $$-\frac{1}{2} (\frac{2}{3}u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}})+C $$ $$\frac{1}{5}u^{\frac{5}{2}}-\frac{1}{3}u^{\frac{3}{2}}+C$$ $$\frac{1}{5}(1-x^2)^{\frac{5}{2}}-\frac{1}{3}(1-x^2)^{\frac{3}{2}}+C$$

Never forget the $+C$ when dealing with indefinite integrals

  • $\begingroup$ Please use an online utility to check your answer I like this one: integral-calculator.com Wolfram alpha also disagrees with you: wolframalpha.com/input/… $\endgroup$ – Anirudh Jul 3 at 3:49
  • $\begingroup$ Bruh I got the same answer as eyeballfrog $\endgroup$ – Anirudh Jul 3 at 4:06
  • $\begingroup$ Sorry. I miscopied your answer into Mathematica and it is correct. $\endgroup$ – robjohn Jul 3 at 4:09
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    $\begingroup$ Never forget the +C when dealing with (indefinite) integrals $\endgroup$ – Hussain-Alqatari Jul 3 at 5:25

To integrate $\displaystyle \int x^3 \sqrt{1 - x^2} dx$, I would most naturally use $u$-substitution with $u = 1 - x^2$ and $du = -2x\hspace{1mm}dx$. Then this is $$\begin{align} \int x^3 \sqrt{1 - x^2} dx &= \frac{-1}{2}\int (-2x\hspace{1mm}dx) (x^2) \sqrt{1 - x^2} \\ &= \frac{-1}{2}\int (1 - u) \sqrt{u} \; du \\ &= \frac{-1}{2} \bigg( \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2}\bigg) + C. \end{align}$$ Substituting back in gives the answer $$ \frac{(1-x^2)^{5/2}}{5} - \frac{(1 - x^2)^{3/2}}{3} + C. $$


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