## ksaimouli 2 years ago antiderivative?

1. ksaimouli

|dw:1354755369490:dw|

2. ksaimouli

plz explain rule ?

3. zepdrix

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. :)

4. ksaimouli

no

5. ksaimouli

:-)

6. zepdrix

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|

7. zepdrix

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,$\large y'=\frac{1}{\sec^2y}\qquad \rightarrow \qquad y'=\cos^2 y$

8. zepdrix

We can extract from the picture what cosine of y is.$\large y'=\left(\frac{1}{\sqrt{1+x^2}}\right)^2 \qquad \rightarrow \qquad y'=\frac{1}{1+x^2}$

9. zepdrix

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

10. zepdrix

What we just did was, We showed that if we start with,$\large y=\tan^{-1}x$Then,$\large y'=\frac{1}{1+x^2}$

11. zepdrix

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

12. ksaimouli

ohhh i got u i memorized these

13. zepdrix

k good idea :3