I need some help with Trigometric substitutions, can someone help me?
Depends... post your question, we'll see :)
\[\int\limits \frac{ 1 }{ \sqrt{1-x ^{2}} }dx\]
Okay, whenever you see a form \[\Large \sqrt{1-x^2}\] particularly in the denominator, think about using the substitution \[\Large x = \sin(\theta)\]
ok
So, it follows that... \[\Large dx = \cos(\theta)d\theta\] yes?
yes
Use that substitution for. \[\Large \sqrt{1-x^2}\] replace x with \(\sin(\theta)\) and tell me what you get...
\[\sqrt{1-\sin(\theta)^{2}}\]
Let's have it as \[\Large \sqrt{1-\sin^2(\theta)}\] I'm sure that's what you meant, now what is this equal to?
Remember that \[\Large \cos^2(\theta) + \sin^2(\theta) = 1\]
ok
So...? What is \[\Large \sqrt{1-\sin^2(\theta)}=\color{red}?\]
\[\cos ^{2}\theta\]
tsk tsk :P \[\Large = \sqrt{1-\sin^2(\theta) }={\LARGE \color{red}{\sqrt{\color{black}{\cos^2(\theta)}}}}\]
Don't forget the radical :P
So anyway... \[\Large \frac1{\sqrt{1-x^2}}=\frac1{\sqrt{1-\sin^2(\theta)}}=\frac1{\cos(\theta)}\] ok?
this would be the final step right?
Almost, we also have to deal with the dx... \[\Large \int \frac1{\cos(\theta)}dx\]but we have an expression for dx in terms of \(d\theta\)
+c
No... remember this? \[\Large x = \sin(\theta)\] so that... \[\Large \frac{dx}{d\theta}=\cos(\theta)\]
\[\int\limits \frac{ 1 }{ \cos \theta}\cos \theta\]
smh :3 \[\Large dx = \cos(\theta)d\theta\] \[\Large \int \frac1{\cos(\theta)}dx = \int \frac1{\cos(\theta)}\cdot \cos(\theta)d\theta\]
I trust you can do it from here?
still not sure about the next steps
Well, the cos cancels out like this... \[\Large \int\frac1{\cancel{\cos(\theta)}}\cancel{\cos(\theta}d\theta=\int d\theta\] and this is possibly the simplest integral there is :P
okay i forgot about crossing out the cos
for the 1 up top would i put that out side of the interval? or because it's a constant do I get rid of it?
Uhh, sure, but 1 times any number is just that number, so the 1 just disappears? -.-
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