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spyros

  • 2 years ago

integral(sec(x)^3)

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  1. experimentX
    • 2 years ago
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    \[ \int \sec^3x dx \text{ or } \int \sec (x^3) dx ??\]

  2. spyros
    • 2 years ago
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    \[\int\limits(\sec(x)^3dx\]

  3. experimentX
    • 2 years ago
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    former or latter??

  4. spyros
    • 2 years ago
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    former

  5. experimentX
    • 2 years ago
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    \[ \sec^3x = \sec^2 \sec x = (1 + \tan^2x) \sec x\]

  6. spyros
    • 2 years ago
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    actually( \[\int\limits{\sqrt(1+(x-1)^2)dx}\]

  7. spyros
    • 2 years ago
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    @experimentX i'll try that one

  8. spyros
    • 2 years ago
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    i came to integral(sec(x)^3dx) from integral(sqrt(1+(x-1)^2)dx

  9. slaaibak
    • 2 years ago
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    \[\int\limits_{}^{} \sec^3(x) = \int\limits_{}^{}\sec x {d \over dx} \tan x = \sec x \tan x - \int\limits \sec (x)\tan^2x \] \[\int\limits \sec^3(x) = \sec x \tan x - \int\limits \sec x (\sec^2(x) - 1) \]

  10. spyros
    • 2 years ago
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    ok, i get it

  11. slaaibak
    • 2 years ago
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    Now take the second part further

  12. experimentX
    • 2 years ago
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    int sqrt(1 + (x-1)^2) dx = (1/4*(2*x-2))*sqrt(2+x^2-2*x)+(1/2)*arcsinh(x-1)

  13. experimentX
    • 2 years ago
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    if case of any confusion plugin integration to wolframalpha and click for show steps

  14. spyros
    • 2 years ago
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    pellet

  15. spyros
    • 2 years ago
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    sh.it

  16. Callisto
    • 2 years ago
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    Just a little try, not sure if it is correct but hope it helps \[\int sec^3xdx\]\[=\int secx(1+tan^2x)dx\]\[=\int secx+secxtan^2xdx\]\[=\int secxdx+\int secxtan^2xdx\]\[=\int \frac{secx(secx+tanx)}{secx+tanx}dx+\int tanxd(secx)\]\[=\int \frac{1}{secx+tanx}d(secx+tanx)+[secxtanx-\int secxd(tanx)]\]\[=\ln |secx+tanx|+[secxtanx-\int secxsec^2xdx] +C\]\[=\ln |secx+tanx|+[secxtanx-\int sec^3xdx] +C\] So, we've got \[\int sec^3xdx=\ln |secx+tanx|+secxtanx-\int sec^3xdx +C\]\[2\int sec^3xdx=\ln |secx+tanx|+secxtanx +C\]\[\int sec^3xdx=\frac{1}{2}(\ln |secx+tanx|+secxtanx) +C\]

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