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anonymous

  • 4 years ago

integrate[sec^3x]

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{?}^{?}\sec ^{3}\theta d \theta\]

  2. anonymous
    • 4 years ago
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    |dw:1327731874453:dw||dw:1327732061041:dw|

  3. anonymous
    • 4 years ago
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    because the reciprocal ratio for sec is 1/cosx

  4. anonymous
    • 4 years ago
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    so then just bring the power to the front

  5. anonymous
    • 4 years ago
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    so

  6. anonymous
    • 4 years ago
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    thanks, never thought about that move;

  7. anonymous
    • 4 years ago
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    |dw:1327732192064:dw|

  8. anonymous
    • 4 years ago
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    also have to know the general power rule wich is

  9. anonymous
    • 4 years ago
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    |dw:1327732252138:dw|

  10. anonymous
    • 4 years ago
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    im not getting the rule.. explain?

  11. TuringTest
    • 4 years ago
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    no that won't quite work above you are going to integration by parts I think here

  12. anonymous
    • 4 years ago
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    it was solved, but idk y he squared the last integral-- and the power rule he stated

  13. TuringTest
    • 4 years ago
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    u=secx du=secxtanx dv=sec^2x u=tanx

  14. TuringTest
    • 4 years ago
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    \[\int \sec^3xdx=\sec x \tan x-\int \sec x\tan^2xdx\]\[=\sec x \tan x-\int \sec x(1-\sec^2x)dx\]\[=\sec x \tan x-\int\sec^3 xdx+\int\sec xdx\]have it from there ?

  15. TuringTest
    • 4 years ago
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    v=tanx above*

  16. anonymous
    • 4 years ago
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    yes, got it; thanks, for the integral of secx, you change to 1/cosx, then multiply by cosx/cos right?

  17. TuringTest
    • 4 years ago
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    \[\int\sec xdx=\int\sec x(\frac{\sec x+\tan x}{\sec x+\tan x})=\int\frac{du}{u}\]

  18. anonymous
    • 4 years ago
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    i c, kk thanks alot man =)

  19. TuringTest
    • 4 years ago
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    anytime

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