I have to find the definite integral of this:

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

So let's start with the indefinite integral:

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


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

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

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

But now I'm stuck...



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

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?

  • 1
    $\begingroup$ 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). $\endgroup$ – Clayton Feb 2 '19 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 $u=\sin\theta$.

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$$F=\dfrac{\sec x\tan x}{(\tan^2x)^{3/2}}=\dfrac{\sec x\tan x}{|\tan^3x|}$$

For $\tan x>0,$

$$F=\dfrac{\cos x}{\sin^2x}=\csc x\cot x=-\dfrac{d(\csc x)}{dx}$$

What if $\tan x<0$

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