Integration by multiple substitutions

Question

I have the following integration. The only instructions are to integrade by substitution.

(again I am sorry but I have no idea how to use MathJax. I am going to have to sit down and figure it out because I am sure it would be easier for everyone.)

$$ \int x^5 \sqrt {x^3 - 8}\,dx $$

So far I have $u = x^3$, $du = 3x^2dx$ giving me:

$$ \frac13 \int u \sqrt {u - 8}\,du $$

And this is where I am stuck.
Any help in the right direction would be fantastic.

Thank you

Answer 1

$$\displaystyle\dfrac{1}{3}\int u\sqrt{u-8}\,du$$ Now, $$\displaystyle\dfrac{1}{3}\left[\int (u-8+8)\sqrt{u-8}\,du\right]$$ $$=\displaystyle\dfrac{1}{3}\left[\int(u-8)^{\frac{3}{2}}du+8\int(u-8)^{\frac{1}{2}}du\right]$$ $Let (u-8)=v$ $\Rightarrow\;du=dv$ $$\Rightarrow\;\displaystyle\dfrac{1}{3}\int\left[v^{\frac{3}{2}}dv+8\int v^{\frac{1}{2}}du\right]$$ Integrate resulting integral $$=\dfrac{1}{3}\left[\dfrac{2(u-8)^{\frac{5}{2}}}{5}+\dfrac{8×2}{3}(u-8)^{\frac{3}{2}}\right]$$ Since you let $u=x^3$

$$=\dfrac{2}{15}(x^3-8)^{\frac{5}{2}}+\dfrac{16}{9}(x^3-8)^{\frac{3}{2}}$$ $$=\dfrac{6}{45}\left[(x^3-8)^\frac{5}{2}\right]+\dfrac{16×5}{9×5}(x^3-8)^{\frac{3}{2}}$$ $$\Rightarrow\;\dfrac{2}{45}(x^3-8)^{\frac{3}{2}}\left[3(x^3-8)+40\right]$$ $$\Rightarrow\;\dfrac{2}{45}(x^3-8)^{\frac{3}{2}}\left[3x^3-24+40\right]$$ $$\Rightarrow\;\dfrac{2}{45}(x^3-8)^{\frac{3}{2}}(3x^3+16)+C$$

Answer 2

Hint. One may write $$ \begin{align} \int u\cdot \sqrt{u-8}\:du&=\int (u-8+8)\cdot \sqrt{u-8}\:du \\&=\int (u-8)^{3/2}\:du+8\int (u-8)^{1/2}\:du \end{align} $$ then just substitute $v=u-8$ to get $$ \int u\cdot \sqrt{u-8}\:du=\int v^{3/2}\:dv+8\int v^{1/2}\:dv $$ the last terms are now easier to evaluate.

Answer 3

In general, when you have an integral of the form

$$ \int R(x,\sqrt{ax+b})dx $$

the substitution $t=\sqrt{ax+b}$ gives

$$ \int R\left(\frac{t^2-b}{a},t\right)\frac{2tdt}{a} $$

If $R$ is a rational function (of two variables), the resulting integral is a rational one.