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12-13-2008   #1 (permalink)
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Default Calculus

I know how this one works, I just thought it was a really awesome question.

∫sec≥(x)dx
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12-14-2008   #2 (permalink)
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∫Sec≥ (x) dx
=∫Sec^2 (x) Sec (x) dx = Sec (x) Tan (x) -∫Sec (x) Tan (x) Tan (x) dx
= Sec (x) Tan (x) - ∫ Sec (x) Tan^2 (x) dx
= Sec (x) Tan (x) - ∫ Sec (x)[Sec^2 (x) - 1] dx

Therefore,

∫ Sec^3 (x) dx = Sec (x) Tan (x) - (∫ Sec (x)[Sec^2 (x) - 1] dx)

Thus,
∫ Sec^3 (x) dx = Sec (x) Tan (x) - ∫ Sec^3 (x) dx + ∫Sec (x) dx

2 ∫ Sec^3 (x) dx = Sec (x) Tan (x) -∫ Sec (x) dx

2 ∫ Sec^3 (x) dx = Sec (x) Tan (x) - ln | Sec (x) + Tan (x) |

Which means:
∫ Sec^3 (x) dx = 1/2 (Sec (x) Tan (x) - ln | Sec (x) + Tan (x) |) + C

There's another method, but I can't recall how to do it on the top of my head.
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Last edited by k0n; 12-14-2008 at 01:00 AM.
 

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