What is the Correct Integral of Inverse Cosecant?

[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)?

 

[certainly]

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

 

[Euler2718]

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

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

 

[certainly]

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

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

 

[Euler2718]

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

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.

 

[certainly]

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.

Hint:- Use trigonometric substitution.

 

[Euler2718]

Hint:- Use trigonometric substitution.

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.

 

[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

 

[tommik]

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