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Mathematics 8 Online
OpenStudy (anonymous):

I need some help with Trigometric substitutions, can someone help me?

terenzreignz (terenzreignz):

Depends... post your question, we'll see :)

OpenStudy (anonymous):

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

terenzreignz (terenzreignz):

Okay, whenever you see a form \[\Large \sqrt{1-x^2}\] particularly in the denominator, think about using the substitution \[\Large x = \sin(\theta)\]

OpenStudy (anonymous):

ok

terenzreignz (terenzreignz):

So, it follows that... \[\Large dx = \cos(\theta)d\theta\] yes?

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

Use that substitution for. \[\Large \sqrt{1-x^2}\] replace x with \(\sin(\theta)\) and tell me what you get...

OpenStudy (anonymous):

\[\sqrt{1-\sin(\theta)^{2}}\]

terenzreignz (terenzreignz):

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?

terenzreignz (terenzreignz):

Remember that \[\Large \cos^2(\theta) + \sin^2(\theta) = 1\]

OpenStudy (anonymous):

ok

terenzreignz (terenzreignz):

So...? What is \[\Large \sqrt{1-\sin^2(\theta)}=\color{red}?\]

OpenStudy (anonymous):

\[\cos ^{2}\theta\]

terenzreignz (terenzreignz):

tsk tsk :P \[\Large = \sqrt{1-\sin^2(\theta) }={\LARGE \color{red}{\sqrt{\color{black}{\cos^2(\theta)}}}}\]

terenzreignz (terenzreignz):

Don't forget the radical :P

terenzreignz (terenzreignz):

So anyway... \[\Large \frac1{\sqrt{1-x^2}}=\frac1{\sqrt{1-\sin^2(\theta)}}=\frac1{\cos(\theta)}\] ok?

OpenStudy (anonymous):

this would be the final step right?

terenzreignz (terenzreignz):

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 \(d\theta\)

OpenStudy (anonymous):

+c

terenzreignz (terenzreignz):

No... remember this? \[\Large x = \sin(\theta)\] so that... \[\Large \frac{dx}{d\theta}=\cos(\theta)\]

OpenStudy (anonymous):

\[\int\limits \frac{ 1 }{ \cos \theta}\cos \theta\]

terenzreignz (terenzreignz):

smh :3 \[\Large dx = \cos(\theta)d\theta\] \[\Large \int \frac1{\cos(\theta)}dx = \int \frac1{\cos(\theta)}\cdot \cos(\theta)d\theta\]

terenzreignz (terenzreignz):

I trust you can do it from here?

OpenStudy (anonymous):

still not sure about the next steps

terenzreignz (terenzreignz):

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

OpenStudy (anonymous):

okay i forgot about crossing out the cos

OpenStudy (anonymous):

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?

terenzreignz (terenzreignz):

Uhh, sure, but 1 times any number is just that number, so the 1 just disappears? -.-

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