How do you set up this Tricky u-sub?

Question

Tricky u-sub. Can you point me in the right direction?

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

$u = 5-2x^2$

$ du = -4x dx$

Obviously, this does not match fully.

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I tried breaking up the $x^3 = x*x^2$ and I continued with:

$u=5-2x^2$

$2x^2 = 5-u$

$x^2=\frac{5-u}{2}$

But, I can't see how to make it all fit.

Am I on the right track?

Answer 1

Am I on the right track?

You're doing fine!

So you have $u=5-2x^2$ which gives $x^2=\frac{5-u}{2}$ and from $\mbox{d}u=-4x\,\mbox{d}x$ you get $x\,\mbox{d}x=-\tfrac{1}{4}\mbox{d}u$.

Now perform the substitution: $$\int{{x^3}\sqrt{5-2x^2}}\,\mbox{d}x = -\frac{1}{4}\int\frac{5-u}{2}\sqrt{u}\,\mbox{d}u= -\frac{1}{8}\int \left(5\sqrt{u}-u^{\tfrac{3}{2}}\right)\,\mbox{d}u = \ldots$$

Answer 2

By your idea we have

• $u = 5-2x^2\implies x^2=\frac{5-u}{2}$
• $ du = -4x dx\implies xdx=-\frac14 du $

then

$$\int{{x^3}\sqrt{5-2x^2}}dx=\int{\frac{5-u}{2}\sqrt{u}}\left(-\frac14\right) du$$

Answer 3

You've done all the hard work. Doing some rearranging, you obtain $$\int x^3 \sqrt{5-2x^2}\, dx = \int -\frac{1}{4} x^2 \sqrt{5-2x^2} \cdot (-4x)\, dx = \int -\frac{1}{4} \cdot \frac{5-u}{2} \sqrt{u}\, du$$ This last integral can be computed as usual by considering powers of $u$.

Answer 4

Another approach:

We can write the integrand as \begin{align*} {x^3}\sqrt{5-2x^2}&=-\tfrac12x(5-2x^2)\sqrt{5-2x^2}+\tfrac52x\sqrt{5-2x^2}&&\text{then}\\[5pt] \int{x^3}\sqrt{5-2x^2}dx&=-\frac12\int x\left(5-2x^2\right)^{3/2}dx+\frac52\int x\left(5-2x^2\right)^{1/2}dx\\[5pt] &=\color{red}{\frac1{20}\left(5-2x^2\right)-\frac5{12}\left(5-2x^2\right)^{3/2}+C} \end{align*}