# $\int x^3$ $\sqrt{1-x^2}\ dx$ [closed]

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

## closed as off-topic by José Carlos Santos, postmortes, max_zorn, Aqua, CesareoJul 3 at 9:17

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• Do you mean "indefinite" instead of "undefined"? – 0XLR Jul 3 at 3:06
• Sorry, I adjusted. – Mycroft Jul 3 at 3:07
• Straightforward integration by parts: $-\frac{1}{15} \left(1-x^2\right)^{3/2} \left(3 x^2+2\right)$. – David G. Stork Jul 3 at 3:09
• Could you explain the steps? – 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 $$\begin{multline} \int x^3\sqrt{1-x^2}dx = -\frac{1}{2}\int(1-u)\sqrt{u}\,du = -\frac{1}{2}\int\left(u^{1/2}-u^{3/2}\right)\,du = \frac{u^{5/2}}{5} - \frac{u^{3/2}}{3}\\ = \frac{(1-x^2)^{5/2}}{5}-\frac{(1-x^2)^{3/2}}{3} = -\frac{2+3x^2}{15}(1-x^2)^{3/2} \end{multline}$$

• (+1) modulo the constant of integration, this is correct. – 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

• 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/… – Anirudh Jul 3 at 3:49
• Bruh I got the same answer as eyeballfrog – Anirudh Jul 3 at 4:06
• Sorry. I miscopied your answer into Mathematica and it is correct. – robjohn Jul 3 at 4:09
• Never forget the +C when dealing with (indefinite) integrals – 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 $$\frac{(1-x^2)^{5/2}}{5} - \frac{(1 - x^2)^{3/2}}{3} + C.$$