# What's wrong with my integral ?

1. Dec 15, 2005

### Josh_K

$$\int \frac{3\sqrt{x^2-16}}{2x}$$

Let $$x = 4\sec\theta$$, $$dx = 4\sec\theta\tan\theta d\theta$$. $$\theta = asec(x/4)$$

$$\frac{3}{2}\int \frac{\sqrt{4^2(\sec^2\theta-1)}}{4\sec\theta}4\sec\theta\tan\theta d\theta$$

$$\frac{3}{2}\int \frac{\sqrt{\tan^2\theta}}{\sec\theta}4\sec\theta\tan\theta d\theta$$

$$\frac{4\times 3}{2}\int \tan^2\theta d\theta$$

as $$tan^2 = sec^2-1$$

$$6\int sec^2 d\theta - 6\int d\theta$$

$$6 \tan\theta - 6\theta$$

As $$\tan\theta = \frac{\sqrt{x^2-16}}{4}$$

$$6 \tan\theta - 6\theta = \frac{3}{2}\sqrt{x^2-16} - 6 arcsec\left(\frac{x}{4}\right) + C$$

I don't know what's wrong but when I plot it on graphmatica and derive it, it doesn't match with the original equation.

2. Dec 15, 2005

### StatusX

That looks right to me. Remember that:

$$\frac{d(\mbox{sec}^{-1} x)}{dx}=\frac{1}{x\sqrt{x^2-1}}$$

3. Dec 15, 2005

### Josh_K

Why then, when I plot it on maple or on graphmatica, it's not the same thing at all. There's a problem somewhere...

4. Dec 15, 2005

### StatusX

Why don't you try differentiating it yourself to see if it gives you back the integrand. Also be careful with the sign of x. Strictly speaking, that should be an absolute value of x in the derivative of arcsecant I gave. And be careful with how arcsecant is defined. Remember it is a multivalued function like arccos, and you have to specify what you want the domain to be. (ie, is arcsec(1)=0, 2pi, 4 pi, ...?)

5. Dec 15, 2005

### GCT

well since you're trying to plot it try using the arctangent alternative, the boundaries may be different.