# Integral of Inverse Cosecant

1. May 10, 2015

### Euler2718

Running into a little trouble when doing this integral by hand:

$$\int arccsc(x) dx$$

$$u = arccsc(x) \implies du = -\frac{1}{x\sqrt{x^{2}-1}} dx$$

$$dv = dx \implies v = x$$

$$\int u dv = uv - \int vdu$$

$$\int arccsc(x) dx = xarccsc(x) - \int x\cdot -\frac{1}{x\sqrt{x^{2}-1}} dx$$

$$\int arccsc(x) dx = xarccsc(x) + \int \frac{1}{\sqrt{x^{2}-1}} dx$$

Now I'm here, and it seems obvious , but this yields the answer to be:

$$\int arccsc(x) dx = xarccsc(x) + \ln\left|\sqrt{x^{2}-1}\right| +C$$

When it's suppose to be (from wikipedia):

$$\int arccsc(x) dx = xarccsc(x) + \ln\left(x+\sqrt{x^{2}-1}\right) +C$$

Where did I go wrong (skipped ahead in the course, so sorry if it's something really basic I'm missing)?

2. May 10, 2015

### certainly

That last integral is wrong.
$\frac{d}{dx}ln|\sqrt{x^2-1}|=\frac{x}{x^2-1}\neq\frac{1}{\sqrt{x^2-1}}$.

Last edited: May 10, 2015
3. May 10, 2015

### Euler2718

Yes, I'm aware. I'm wondering where I messed up in the integration by parts.

4. May 10, 2015

### certainly

The integration by parts is right.
However, $\int \frac{1}{\sqrt{x^2-1}} dx \neq ln|\sqrt{x^2-1}|$.

5. May 10, 2015

### Euler2718

hm..How exactly does $\int \frac{1}{\sqrt{x^2-1}} dx$ boil down to $\ln \left( x + \sqrt{x^{2}-1} \right)$ then? I'm a bit lost now.

6. May 10, 2015

### certainly

Hint:- Use trigonometric substitution.

7. May 10, 2015

### Euler2718

Thank you! I never heard of trigonometric substitution lol (like I said, I skipped ahead). Did a secant substitution, and got the real answer. Thank you.

8. May 11, 2015

### tommik

if you do not want to use trigonometric subst or doing boring calculation, you can multiply and divide $\left( 1/ \sqrt{x^{2}-1} \right)$ by

$\left( x + \sqrt{x^{2}-1} \right)$

in this way the result is immediate! In fact you find

$\left( 1 + x/\sqrt{x^{2}-1} \right)/ \left( x + \sqrt{x^{2}-1} \right)$

that is f'/f , so the result is $\log \left( x + \sqrt{x^{2}-1} \right)$

panta rei

Last edited: May 11, 2015
9. May 12, 2015

### tommik

..and obviously with the absolute value bars....