# Thread: Another trig sub integration

1. ## Another trig sub integration

Hey

Im stuck on the last step of another trig sub again...

Heres the problem: integrate 1/((x^2)*(sqrt(4x^2-9))

I've sub'd x=3/2sec(theta)

I get to the point after integrating ..(2sin(theta))/9 and have to sub back in for theta

I realise theta=sec^-1(2x/3) but do not understand how i can use 2(sin(sec^-1(2x/3))/9 to finish the problem.

thanks

2. Originally Posted by Sam1111
Hey

Im stuck on the last step of another trig sub again...

Heres the problem: integrate 1/((x^2)*(sqrt(4x^2-9))

I've sub'd x=3/2sec(theta)

I get to the point after integrating ..(2sin(theta))/9 and have to sub back in for theta

I realise theta=sec^-1(2x/3) but do not understand how i can use 2(sin(sec^-1(2x/3))/9 to finish the problem.

Create a right angled triangle where an interior angle is $\displaystyle \theta$ and the $\displaystyle \sec \theta = \frac{2x}{3}$ or $\displaystyle \cos \theta = \frac{3}{2x}$. So the adjacent is $\displaystyle 3$ and the hypotenuse is $\displaystyle 2x$. The pythagorean thm gives the opposite as $\displaystyle \sqrt{4x^2-9}$ and from here you can pick of the $\displaystyle \sin \theta$.