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Math Help - Integral of (tanx)^71 * (secx)^4

  1. #1
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    Integral of (tanx)^71 * (secx)^4

    \int tan^71x sec^4x

    \int tan^70x sec^3x * tanx secx


    I'm stuck after this part
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  2. #2
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    I wouldn't do that.

    My first impression is to take two of the secants and turn them into tangents.

    After that, it is a rather obvious substitution u = tan(x) and you're done.
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  3. #3
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    Hello, circuscircus!

    TKHunny has the best idea . . .


    \int \left(\tan^{71}\!x\right)\left(\sec^4\!x\right)\,d  x

    We have: . \left(\tan^{71}\!x\right)\left(\sec^2\!x\right)\le  ft(\sec^2\!x\right) \;=\;\left(\tan^{71}\!x\right)\left(\tan^2\!x+1\ri  ght)\left(\sec^2\!x\right)

    . . Then: . \int \left(\tan^{73}\!x + \tan^{71}\!x\right)(\sec^2\!x)\,dx

    Let: u \:= \:\tan x\quad\Rightarrow\quad du \:=\:\sec^2\!x\,dx

    Substitute: . \int \left(u^{73} + u^{71}\right)\,du . . . . Got it?

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