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anonymous
 3 years ago
Area under y = arctan(x) between x = 1 and x = 0
anonymous
 3 years ago
Area under y = arctan(x) between x = 1 and x = 0

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand how to integrate arctan(x) by hand

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think i remember it having something to do with integration by parts...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We have to somehow use the tan function though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It says use the fact that y=artan(x) is the inverse of y = tan(x) to calculate the area between x = 1 and x =0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We can also graph it for a hint

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But we haven't learned integration by parts yet

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.2@baddinlol You have to use integration by parts here, \[\int\limits \tan ^{1}(x) \, dx\] \[u=\tan^{1}x~~~~~dv=dx \\ \\ du=\frac{1}{1+x^2} ~~~~~v=x\] \[x \tan ^{1}x\int\limits \frac{x}{1+x^2} \, dx\] Then continue...
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