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ksaimouli
 4 years ago
antiderivative?
ksaimouli
 4 years ago
antiderivative?

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ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354755369490:dw

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Hmm 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. :)

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1We 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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Taking 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,\[\large y'=\frac{1}{\sec^2y}\qquad \rightarrow \qquad y'=\cos^2 y\]

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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1It'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

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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1So 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

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh i got u i memorized these
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