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ksaimouli
antiderivative?
|dw:1354755369490:dw|
Hmm this one is a little tricky. Have you been introduced to Trig Substitutions yet? I mean, this is a very common anti-derivative 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. :)
We can show that the DERIVATIVE of arctangent = 1/(1+x^2) From that information, we'll be able to determine that the ANTI-DERIVATIVE 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|
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,\[\large y'=\frac{1}{\sec^2y}\qquad \rightarrow \qquad y'=\cos^2 y\]
We can extract from the picture what cosine of y is.\[\large y'=\left(\frac{1}{\sqrt{1+x^2}}\right)^2 \qquad \rightarrow \qquad y'=\frac{1}{1+x^2}\]
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
What we just did was, We showed that if we start with,\[\large y=\tan^{-1}x\]Then,\[\large y'=\frac{1}{1+x^2}\]
So if we STARTED with,\[\large \frac{1}{1+x^2}\]And wanted to know it's anti-derivative, then we just go back that direction :D
ohhh i got u i memorized these