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ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
dw:1354755369490:dw
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
plz explain rule ?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Hmm this one is a little tricky. Have you been introduced to Trig Substitutions yet? I mean, this is a very common antiderivative that will come up over and over, so it's worth your time to memorize it. But if you really want to understand where it's coming from, then that requires a little more work. :)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We can show that the DERIVATIVE of arctangent = 1/(1+x^2) From that information, we'll be able to determine that the ANTIDERIVATIVE of 1/(1+x^2) =arctangent + c To be able to do it in the other direction requires a Trig Sub which is a bit tricky ^^ Start here,\[\large \text{Let} \quad y=\tan^{1} x\]Recall that we can rewrite this as the TANGENT by switching the arguments.\[\large x=\tan y\]Let's label this on a triangle so we can reference it.dw:1354755709197:dw
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Taking the derivative of our function with respect to X gives us,\[\large x=\tan y \qquad \rightarrow \qquad 1=y'\cdot \sec^2y \]Solving for y' gives us,\[\large y'=\frac{1}{\sec^2y}\qquad \rightarrow \qquad y'=\cos^2 y\]
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We can extract from the picture what cosine of y is.\[\large y'=\left(\frac{1}{\sqrt{1+x^2}}\right)^2 \qquad \rightarrow \qquad y'=\frac{1}{1+x^2}\]
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
It's a bit of a tricky one to get through, there might be an easier way to do it. But this is the way that I've remembered at least. :\ Confused anywhere in there? There a bunch of crazy stuff going on I know :D
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
What we just did was, We showed that if we start with,\[\large y=\tan^{1}x\]Then,\[\large y'=\frac{1}{1+x^2}\]
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So if we STARTED with,\[\large \frac{1}{1+x^2}\]And wanted to know it's antiderivative, then we just go back that direction :D
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
ohhh i got u i memorized these
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
k good idea :3
 2 years ago
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