# Thread: Need help for an integral

1. ## Need help for an integral

I didn't recall
but I don't think my result is that strange or difficult to compute.

The integral is $\int \frac{dx}{x\sqrt{1-x^2}}$. I'm sorry if I posted it before, but I don't think so (even in that case, I would now solve it differently).
I made the substitution $x=\sin(\theta)$ and I could reach that the integral's worth $\int \cot (\theta )\cdot \sec(\theta)d\theta$. I don't know how to solve this one...

2. $\int \frac{dx}{x\sqrt{1-x^2}} = \int \frac{x~dx}{x^2\sqrt{1-x^2}}$

Let $u = \sqrt{1-x^2}$, then $u^2 = 1-x^2$ and $x~dx = -u~du$.

$\int \frac{-\not u~du}{(1-u^2)\not u} = -\int \frac{1}{1-u^2} ~du$. I think you can finish it from here.

3. $\int\frac{dx}{x\sqrt{1-x^2}}$

Let $x=\sin(\theta)$

So

$dx=\cos(\theta)~d\theta$

So we have

$\int\frac{\cos(\theta)}{\sin(\theta)\sqrt{1-\sin^2(\theta)}}~d\theta$

$=\int\csc(\theta)$

$=-\ln|\csc(\theta)+\cot(\theta)|=\ln\left|\tan\left( \frac{x}{2}\right)\right|$

$\underbrace{=}_{\text{backsub}}\ln\left|\tan\left( \frac{\arcsin(x)}{2}\right)\right|$

4. Mathstud28, you wrote
$=-\ln|\csc(\theta)+\cot(\theta)|=\ln\left|\tan\left( \frac{x}{2}\right)\right|$
, shouldn't it be $=-\ln|\csc(\theta)+\cot(\theta)|=\ln\left|\tan\left( \frac{\theta}{2}\right)\right|$? Also, I understand everything you did but not this step. Could you detail a bit more this step? I also guess that $=\int\csc(\theta)$ is a value I MUST know, right?

5. Originally Posted by arbolis
Mathstud28, you wrote , shouldn't it be $=-\ln|\csc(\theta)+\cot(\theta)|=\ln\left|\tan\left( \frac{\theta}{2}\right)\right|$? Also, I understand everything you did but not this step. Could you detail a bit more this step? I also guess that $=\int\csc(\theta)$ is a value I MUST know, right?
Yeah it should be theta not x

And this is one you should know.

6. Thanks Mathstud28! I could reach $=-\ln|\csc(\theta)+\cot(\theta)|=\ln\left|\tan\left( \frac{\theta}{2}\right)\right|$
And wingless,

Let , then and .

. I think you can finish it from here.
Very nice way to solve it! The problem wanted a trig or hyperbolic sub, but I should train myself to any way, so I'll try it either. Any prob I get, I post here.