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arctans and integrals and proof and all that jazz watch

1. given that y = arctan 3x, I can easily prove that dy/dx = 3 / (1 + 9x^2), but then, given t hat dy/dx = 3 / (1 + 9x^2), it's IMPOSSIBLE to show that

integral [limits (root3 /3) and 0] 6x arctan 3x dx = 1/9 (4 pi - 3root3).

Or is it?
Any suggestions?
I don't have a clue, and this horrid exam is on Monday... wahh
2. (Original post by lesser weevil)
given that y = arctan 3x, I can easily prove that dy/dx = 3 / (1 + 9x^2), but then, given t hat dy/dx = 3 / (1 + 9x^2), it's IMPOSSIBLE to show that

integral [limits (root3 /3) and 0] 6x arctan 3x dx = 1/9 (4 pi - 3root3).

Or is it?
Any suggestions?
I don't have a clue, and this horrid exam is on Monday... wahh
∫ [0 to 1/√3] 6x arctan 3x dx
= 3x²atan(3x)[0 to 1/√3] - 3∫ 3x²/(1+9x²) dx (using intgeration by parts)
= {((3/3)atan(√3)) - (0)} - {∫ 1 - 1/(1+9x²) dx}
= π/3 - { x - (1/3)atan(3x)[0 to 1/√3]}
= π/3 - {(1/√3 - (1/3).(π/3)) - (0)}
= π/3 - 1/√3 + π/9
= 4π/9 - 1/√3
= (1/9)(4π - 3√3)
==============
3. As attached
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4. Hehe. I often say "and all that jazz" after things. The other is "good enough for jazz" which comes from the difference between tuning up in an orchestra and in a jazz band.
5. (Original post by SsEe)
Hehe. I often say "and all that jazz" after things. The other is "good enough for jazz" which comes from the difference between tuning up in an orchestra and in a jazz band.

Well I say it from the song "All that Jazz". Some old song... bit weird, but funny enuf.

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