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anonymous

  • 5 years ago

integrate tan to the power 5X

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  1. anonymous
    • 5 years ago
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    \[\tan(x)^{5x}\] Is that the expression?

  2. anonymous
    • 5 years ago
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    variable in the exponent, consider logarthmic manipulation: integrate 5x ln tan theta

  3. anonymous
    • 5 years ago
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    Yeah, you didn't say tan of what

  4. anonymous
    • 5 years ago
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    I'd guess (in the absence of any variable except x) that it is: \[\tan^{5}x \] But of course, I could be wrong.

  5. anonymous
    • 5 years ago
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    Where is oneprince?

  6. anonymous
    • 5 years ago
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    The prince has stepped in his concubine to satisfy one of his maidens

  7. anonymous
    • 5 years ago
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    INweton integrate it.that is the correct question

  8. anonymous
    • 5 years ago
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    Re-write it as \[\tan x(\sec^2x -1)(\sec^2x -1) \] Expand and do it term-by-term. Reduction formulae seems unnecessary, and I can;t see anything quicker right now :@.

  9. anonymous
    • 5 years ago
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    You simply let tan x be u and integrate u^5

  10. anonymous
    • 5 years ago
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    INewton continue

  11. anonymous
    • 5 years ago
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    OK, I'll do a little more. That expand to \[\tan x \sec^4x - 2\tan x\ \sec^2x + \tan x \] And note that \[\frac{\mathbb{d}}{\mathbb{d}x}\frac{1}{n}\ \sec^nx =\tan x\ \sec^nx \] Again, probably something far quicker.

  12. anonymous
    • 5 years ago
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    Chaguanas, that doesn't work well because you'll end up with a factor \(cos^2x\) from the substitution of dx.

  13. anonymous
    • 5 years ago
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    I think it does work (sec^2x = 1 + tan^2x) but it's not MUCH nicer. Again, pretty tired so may be wrong.

  14. anonymous
    • 5 years ago
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    Actually, looking at what comes out I'd wager not nicer full stop.

  15. anonymous
    • 5 years ago
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    Alternative method: tables: integration tan^u du=(1/(n-1)) tan^(n-1) u - integral tan^(n-2) u du

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