Integral of $\int x^5 \cdot \sqrt{2 - x^3}dx$

Question

I have been working on this integral and cannot seem to notice an error I am making. Thanks for any advice in advance!

For the following integral $$ \int x^5\sqrt{2 - x^3}\,dx $$ I considered the $u$-substitution as such $$ u = x^3, \quad du = 3x^2\,dx $$ Then I rewrote the integral as follows: $$ \frac{1}{3}\int u\sqrt{2 - u}\,du $$ This I integrated by parts as follows \begin{align} \frac{1}{3}\int u\sqrt{2 - u}\, du &= \frac{1}{3}\left( -\frac{2u\left(2-u\right)^{\frac{3}{2}}}{3} + \frac{2}{3}\int\left(2-u\right)^{\frac{3}{2}}\, dx \right) \\ &= -\frac{2x^3\left( 2-x^3\right)^{\frac{3}{2}}}{9} - \frac{4\left(2-x^3\right)^{\frac{5}{2}}}{45} + C \end{align} where $C$ is the integrating constant. The issue is that WolframAlpha obtains a different answer as such: $$ \frac{2}{-9} (2-x^3)^{2/3} \left(x^3+\frac{2}{3}\right) + D $$ I was expecting it to only differ by the integrating constant but when I graphed them I realised they are not the same. Thanks for any insight into my mistake!

Update: The mistake was found to be in the WolframAlpha interpreter as it was interpreted in the wrong manner, despite the fact that the only different thing I used was a different variable. In conclusion, be sure to put in precise and clear input!

Answer 1

It's easier to avoid the integration by parts entirely by choosing instead $$u = 2-x^3, \quad du = -3x^2 \, dx, \quad x^3 = 2-u,$$ hence $$\begin{align} \int x^5 \sqrt{2-x^3} \, dx &= -\frac{1}{3} \int x^3 \sqrt{2-x^3} (-3x^2) \, dx \\ &= -\frac{1}{3} \int (2-u) \sqrt{u} \, du \\ &= -\frac{1}{3} \int 2 u^{1/2} - u^{3/2} \, du \\ &= -\frac{1}{3} \left( \frac{4}{3} u^{3/2} - \frac{2}{5} u^{5/2} \right) + C \\ &= \frac{2}{15} u^{5/2} - \frac{4}{9} u^{3/2} + C \\ &= \frac{2}{15} (2-x^3)^{5/2} - \frac{4}{9} (2-x^3)^{3/2} + C \\ &= -\frac{2}{45}(4+3x^3)(2-x^3)^{3/2} + C. \end{align}$$

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Your antiderivative is equivalent to the above result and therefore is correct: factorization yields $$-(2-x^3)^{3/2}\left(\frac{2}{9} x^3 + \frac{4}{45}(2-x^3)\right) = -\frac{2}{45}(2-x^3)^{3/2} (4+3x^3).$$

However, the result that you claim you obtained from Wolfram Alpha is incorrect. Differentiation of this expression yields $$\frac{2 x^2 \left(15 x^3-14\right)}{27 \sqrt[3]{2-x^3}}$$ which is not the original integrand.

Answer 2

$$ \begin{aligned} \int x^5 \sqrt{2-x^3} d x = & -\frac{2}{9} \int x^3 d\left(2-x^3\right)^{\frac{3}{2}} \\ = & -\frac{2}{9} x^3\left(2-x^3\right)^{\frac{3}{2}}+\frac{2}{3} \int x^2\left(2-x^3\right)^{\frac{3}{2}} d x \\ = & -\frac{2}{9} x^3\left(2-x^3\right)^{\frac{3}{2}}-\frac{4}{45}\left(2-x^3\right)^{\frac{5}{2}}+C \\ = & -\frac{2\left(2-x^3\right)^{\frac{3}{2}}}{45}\left[5 x^3+2\left(2-x^3\right)\right]+C \\ = & -\frac{2\left(2-x^3\right)^{\frac{3}{2}}}{45}\left(3 x^3+4\right)+C \end{aligned} $$