# How to integrate this using trig substitution?

I have to find the definite integral of this:

$$\int_2^3 \frac{dx}{(x^2-1)^{\frac{3}{2}}}$$

so $$x = \sec \theta$$ so $$dx = \sec \theta \tan \theta d \theta$$

So

$$\frac{\sec{x} \tan{x}}{(\sec^2{x}-1)^{\frac{3}{2}}}$$

$$= \int \frac{\sec{x} \tan{x}}{\tan x^{\frac{3}{2}}}$$

$$= \int \frac{\sec{x}\tan{x}}{\tan{x}^{\frac{1}{2}}}$$

But now I'm stuck...

EDIT

Unstuck:

$$\int \frac{cos \theta}{sin^2 \theta}$$

Let's use $$u = sin \theta$$

$$\int \frac{1}{u^2} du$$

$$\frac{u^-1}{-1} + c$$

$$- \frac{1}{sin \theta} + c$$

So given that $$x = sec \theta$$:

$$- \frac{1}{\frac{\sqrt{x^2-1}}{x}}$$

$$- \frac{x}{\sqrt{x^2-1}}$$

How does that look?

• I think you need to check on your trigonometric identities again. You made a mistake in writing one of them down (perhaps more, but one mistake is glaring). – Clayton Feb 2 at 19:15

You made a slight mistake: since $$\sec^2\theta-1=\tan^2\theta$$ you should have $$\int\frac{\sec\theta\tan\theta d\theta}{\tan^3\theta}=\int\frac{\cos\theta d\theta}{\sin^2\theta}$$, now use $$u=\sin\theta$$.
$$F=\dfrac{\sec x\tan x}{(\tan^2x)^{3/2}}=\dfrac{\sec x\tan x}{|\tan^3x|}$$
For $$\tan x>0,$$
$$F=\dfrac{\cos x}{\sin^2x}=\csc x\cot x=-\dfrac{d(\csc x)}{dx}$$
What if $$\tan x<0$$