Calculate $f(x)=\int_1^x \sqrt{t(3t-2)} \, dt$

Question

I am struggling with this integral calculation.

Calculate:

$$f(x)=\int_1^x \sqrt{t(3t-2)} \, dt$$

for $x\geq 0$

Answer 1

Hint:

Let $u = 3t - 2\implies\mathrm dt = 3\,\mathrm du$. Therefore,

$$\int\sqrt{t(3t - 2)}\,\mathrm dt\equiv\frac1{3^{3/2}}\int\sqrt{u^2 + 2u}\,\mathrm du = \int\sqrt{(u + 1)^2 - 1}\,\mathrm du$$

Let $u + 1 = \sec (v)\implies\mathrm du = \sec(v)\tan(v)\,\mathrm dv$. Therefore, $$\begin{align}\int\sqrt{(u + 1)^2 - 1}\,\mathrm du&\equiv\int\sqrt{\sec^2(v) - 1}\sec(v)\tan(v)\,\mathrm dv \\ &= \int\sec(v)\tan^2(v)\,\mathrm dv \\ &= \int\sec^3(v) - \sec(v)\,\mathrm dv\end{align}$$

From here, linearity is your friend. Notice that the reduction formula for $$\int\sec^nx\,\mathrm dx$$ is well known. You fill find at the end that the integral is only defined for $x\ge 0$.

Answer 2

Complete square

$$f(x)=\int_1^x \sqrt{3t^2-2t} \> dt = \frac1{\sqrt3} \int_1^x\sqrt{(3t-1)^2-1}\>dt$$

and let $3t-1=\cosh u$. Then

\begin{align} f(x)=&\ \frac1{3\sqrt3}\int_{\cosh^{-1}2}^{\cosh^{-1}(3x-1)}\sinh^2 u \, du\\ =& \ \frac1{6\sqrt3}\left(-u+\sinh u\cosh u\right) \bigg|_{\cosh^{-1}2}^{\cosh^{-1}(3x-1)}\\ =& \ \frac1{6\sqrt3}\left(\cosh^{-1}2-\cosh^{-1}(3x-1)+\sqrt3(3x-1)\sqrt{3x^2-2x} -2\sqrt3 \right) \end{align}