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Mathematics 19 Online
OpenStudy (ksaimouli):

antiderivative?

OpenStudy (ksaimouli):

|dw:1354755369490:dw|

OpenStudy (ksaimouli):

plz explain rule ?

zepdrix (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. :)

OpenStudy (ksaimouli):

no

OpenStudy (ksaimouli):

:-)

zepdrix (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|

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