# Evaluate this inverse trig integral PLZZ!!!?

• December 3rd 2008, 07:15 PM
HMV
Evaluate this inverse trig integral PLZZ!!!?
f(x)=1/(x(9x^2-4)^(1/2))
• December 3rd 2008, 08:25 PM
Mathstud28
Quote:

Originally Posted by HMV
f(x)=1/(x(9x^2-4)^(1/2))

You have $\int\frac{dx}{x\sqrt{9x^2-4}}$

Instead of making a trig sub try using a reciporcal substitution first. Let $x=\frac{1}{\varphi}$, so our integral becomes

\begin{aligned}\int\frac{dx}{x\sqrt{9x^2-4}}&\stackrel{x=\frac{1}{\varphi}}{\longmapsto}\in t\frac{\frac{-1}{\varphi^2}}{\frac{1}{\varphi}\sqrt{\frac{9}{\va rphi^2}-4}}\\
&=-\int\frac{d\varphi}{\sqrt{9-4\varphi^2}}\end{aligned}

Now this is a much more common case of letting $2\varphi=3\sin(\phi)$

See how that goes.

You could also have in the initial step let either $3x=2\sec(\phi)$ or $3x=2\cosh(\phi)$