Integral Trouble? Get Help Here! | Cliowa

[cliowa]

Dear community

I'm trying to get a grip on this integral:
$$\int \frac{\sqrt{1-x}}{\sqrt{x}-1} dx$$.
I tried substituting $$x=\sin^{2}(u)$$, which leaves me (standing) with
$$\int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du$$.

But I just can't solve it, no matter which way I try.
I would be thankful for every kind of hint/explanation.
Best regards...Cliowa

 

[arildno]

Well,.when nothing else works, make the s=tan(u/2) substitution:
$$\cos(u)=\cos^{2}\frac{u}{2}-\sin^{2}\frac{u}{2}=\cos^{2}(\frac{u}{2})(1-s^{2})=\frac{(1-s^{2})}{\sec^{2}\frac{u}{2}}=\frac{(1-s^{2})}{1+s^{2}}$$
$$\frac{\sin(u)}{\sin(u)-1}=\frac{2\sin\frac{u}{2}\cos\frac{u}{2}}{2\sin\frac{u}{2}\cos\frac{u}{2}-\cos^{2}\frac{u}{2}-\sin^{2}\frac{u}{2}}=\frac{2s}{2s-s^{2}-1}=-\frac{2s}{(s-1)^{2}}$$

$$u=2\arctan(s)\to{du}=\frac{2ds}{1+s^{2}}$$

Collecting, we get to evaluate the integral:
$$-\int\frac{4s(1+s)^{2}}{(1+s^{2})^{3}}ds$$

 

[TD]

Before introducing trigonometry, I'd substitute $$y = \sqrt{1-x}$$.

 

[VietDao29]

...I tried substituting $$x=\sin^{2}(u)$$, which leaves me (standing) with
$$\int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du$$...

You are forgetting a factor of 2. It should read:
$$2 \int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du$$ instead.
Now by using the Pythagorean Theorem, we have:
cos²x = 1 - sin²x = -(sin x - 1) (sin x + 1)
So you'll have:
$$2 \int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du = -2 \int \sin u (\sin u + 1) du$$.
You can go from here, right? :)

 

[arildno]

That was a bit simpler than mine, VietDao..

 

[cliowa]

Thank you very much, arildno, VietDao29 for your great help.
I figured out the rest on my own.
And, although s=tan(u/2) is a bit more complicated I think it's always good to have to ways to go.
Best regards...Cliowa