Question

Integrate.

Answer 1

\[\int \frac{dx}{(1 - x)\sqrt{1 - x^2}}\]

Answer 2

I tried \(t^2 := 1 - x\).

Answer 3

@hartnn @satellite73

Answer 4

\[1 - x = t^2\]\[t = \sqrt{1 - x}\]\[x = -(t^2 - 1) \Rightarrow \sqrt{1 + x} = \sqrt{2 - t^2}\]\[dx = -2t dt\]Seems like a tedious approach.

Answer 5

So our new integral would be\[- \int \dfrac{2t}{t^3 \sqrt{2 - t^2}} dt\]

Answer 6

So\[-2 \int \frac{dt}{t^2 \sqrt{2 - t^2}}\]

Answer 7

split the integral

Answer 8

1= (1-x) +x

Answer 9

Seems like a better approach.\[\int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{x}{(1 - x) \sqrt{1 - x^2}}\]

Answer 10

Forgot the dx in the end. Oops.

What does the latter integral look like? Is that where the substitution comes into play?

Answer 11

the 2nd integral seems equally complex as the integral in question...seems splitting didn't help much...

Answer 12

In the integral I got, what if I write the numerator as \(\frac{1}{2}\left(t^2 + 2 - t^2 \right)\)?

Answer 13

Hmm

Answer 14

Trig sub should do the trick, \(x=\sin u\). You end up with something like
\[\int\frac{du}{1-\sin u}\]
which can be computed with a half-angle tangent sub, or you can use identities to your advantage.

Answer 15

The identity route is probably less tedious.
\[\frac{1}{1-\sin u}=\frac{1+\sin u}{\cos^2u}=\sec^2u+\sec u\tan u\]