Question

Integral dx/sqrt(1-2x-x^2) by trigonometric substitution

Answer 1

First off, I think it'd be a good idea to complete the square.

Answer 2

2-(x+1)^2 am i right?

Answer 3

You got it, now make the substitution\[ (x+1)=\sqrt{2}\sin(\theta)\] because that will allow you to get rid of the square root on the denominator. I can explain further if you like, it's sort of weird, but just go through it like a u-substitution now where you find dx.

Answer 4

it's not like... integral dx/sqrt 2-(x+1)^2...
so x=sen-1 and dx=cos ???

Answer 5

See, by making this particular substitution, you'll have:

\[\sqrt{2-2\sin^2(\theta)}\]

which if you use this idenitity\[2\sin^2 \theta +2 \cos ^2 \theta =2\]
it means that you'll have \[\sqrt{2\cos^2\theta}=\sqrt2 \cos \theta\] in the denominator.

Answer 6

\[dx=\sqrt{2} \cos \theta \] ???

Answer 7

Yes, almost, just forgot the last part:
\[dx=\sqrt2 \cos \theta d \theta\]

Now just plug them in and you get a pretty nice integral. =)

Answer 8

ohh thanks i really appreciate :)