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Math Final
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May 13th, 2006 19:54 #1 |
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francis |
Does anyone know how to get the reduction formula for sec^n x (calculus II). I think my teacher said integration by parts. Thanks a lot. ______________________ Islam is just the ramblings of some dillusioned Arab madman |
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May 13th, 2006 21:46 #2 |
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robert_ak |
The reduction formula is: To get that, use integration by parts with: u = sec^(n-2) x du = (n-2)*sec^(n-3) x*sec(x) * tan(x) dv = sec^2 x v = tan x We use these values to get dv = sec^2 x because it's easy to integrate (but we would have used sin^(n-1) x for the reduction formula of sin x because sin x is easy to integrate). Then, as usual with the reduction formulas, you will get the integral back when you simplify by using tan^2 x = sec^2 x - 1 (which you get by dividing everything by cos^2 x in sin^2 x + cos^2 x = 1). I hope this helps! 
Last edited by robert_ak on May 13th, 2006 21:46; edited 1 time in total. ______________________ hai |
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May 14th, 2006 12:05 #3 |
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francis |
oooooooooooooooooh i tried sec(n-1) x and it seemed complicated, thanks so much, 
______________________ Islam is just the ramblings of some dillusioned Arab madman |
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May 15th, 2006 00:31 #4 |
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tomzx user |
generally, you'd have to use the loop technique to integrate sec^n, but some "formulas" exists, but no one uses them anyway, except if you have the right to use a book in a math exam which would be stupid in my opinion since looping is quite easy.
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