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baddinlol

  • 2 years ago

Area under y = arctan(x) between x = 1 and x = 0

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  1. baddinlol
    • 2 years ago
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    I don't understand how to integrate arctan(x) by hand

  2. lgbasallote
    • 2 years ago
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    i think i remember it having something to do with integration by parts...

  3. mukushla
    • 2 years ago
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    yep

  4. baddinlol
    • 2 years ago
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    We have to somehow use the tan function though

  5. baddinlol
    • 2 years ago
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    It 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

  6. baddinlol
    • 2 years ago
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    Any ideas?

  7. baddinlol
    • 2 years ago
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    We can also graph it for a hint

  8. baddinlol
    • 2 years ago
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    But we haven't learned integration by parts yet

  9. .Sam.
    • 2 years ago
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    @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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