Question

(1-x^2)^(-1/2)dx

Answer 1

\[x=\sin \theta;\int\limits\limits\frac{\cos \theta d \theta }{ \sqrt{1-\sin^2 \theta} }=\int\limits\limits\frac{ \cos \theta d \theta }{ \sqrt{\cos^2 \theta} }\]\[x=itan \theta;\int\limits\limits\frac{i \sec^2 \theta d \theta }{ \sqrt{1+\tan^2 \theta} }=i\int\limits\limits\frac{ \sec^2 \theta d \theta }{ \sqrt{\sec^2 \theta} }\]

Answer 2

So is this allowed?

Answer 3

\[i*\ln|\sec \theta+\tan \theta|+C = i*\ln|\sqrt{x^2-1}-ix|+C \]

Answer 4

I mean, that's not the point, I just wonder if that's allowed lol.

Answer 5

I'm confused, how are you getting a square under the radical?

Answer 6

Uhhh this way! edits...

Answer 7

Ah ok c:

Answer 8

So if we set them equal to each other, that's fair to do, right? Is this a true statement?

\[i∗\ln|\sqrt{1-x^2}−ix|= \sin^{-1}x+C\]

Answer 9

Now let's manipulate this with a little algebra...

\[\frac{ 1 }{ \sqrt{1-x^2}−ix }= e^{isin^{-1}x}\]

Note, I dropped the +C, but I think it's fair cause this surprised me and I'm not sure if this means anything or what but... euler's identity seems to magically work here.

\[\frac{ 1 }{ \sqrt{1-x^2}−ix }= \cos(\sin^{-1}x)+i \sin(\sin^{-1}x)=\sqrt{1-x^2}+x\]

Answer 10

\[\frac{ 1 }{ \sqrt{1-x^2}−ix }=\sqrt{1-x^2}+ix\] that's what the last thing should say, I forgot the "i" on x.

Answer 11

That's basically just showing multiplying complex conjugates =1. Fancy huh?! What's it mean though lol

Answer 12

\[\large i\cdot \ln|\sec \theta + \tan \theta|\qquad = \qquad i \cdot \ln\left|\frac{\sqrt{x^2-1}}{i}+\frac{x}{i}\right|\]I'm confused by the inside of your logarithm. Did I setup my triangle correctly?