U-substitution of 2x in trigonometric substitution

Question

Find

$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$

The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?

Answer 1

When $u = 2x$, $du = 2dx$, $$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx = \int^{3\sqrt{3}}_0\frac{u^3/8}{(u^2+9)^{3/2}}\frac{du}{2}, $$ this allows us to do further substitution $u = 3 \tan t$ to get rid of the root sign in the denominator. If you use $u = x^3$, $du = 3x^2 dx$, $$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx = \int^{81\sqrt{3}/8}_0\frac{\sqrt[3]{u}/3}{(u^{2/3}+9)^{3/2}}du, $$ which is more complicated then the first equality as this involves the cubic root.

Answer 2

The hint: $$\frac{x^3}{\sqrt{(4x^2+9)^3}}=\frac{x^3+\frac{9}{4}x-\frac{9}{4}x}{\sqrt{(4x^2+9)^3}}=\frac{x}{4\sqrt{4x^2+9}}-\frac{9x}{4\sqrt{(4x^2+9)^3}}.$$ Can you end it now?

Answer 3

Alternatively, change: $$4x^2+9=u \Rightarrow x^2=\frac{u-9}{4}; xdx=\frac{du}{8};$$ $$\int_{9}^{36} \frac{\frac{u-9}{4}\cdot \frac{du}{8}}{u^{3/2}}=\frac{1}{32}\int_{9}^{36} \frac{1}{u^{1/2}}du-\frac{9}{32}\int_{9}^{36} \frac{1}{u^{3/2}}du=\cdots$$