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Math Help - Integrate - Using 2 Different Methods

  1. #1
    Member VitaX's Avatar
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    Integrate - Using 2 Different Methods

    \int tan^5(x)sec^6(x) dx

    I'm told to write them as even exponents so \rightarrow \int (tan^2x)^2(tanx)(sec^2x)^3 dx

    I'm to use 2 methods when solving this, here's the first one:

    \int (tan^2x)^2(tanx)(tan^2x + 1)^2(sec^2x) dx

    u = tanx \rightarrow du = sec^2x dx

    = \int (u^2)2(u)(u^2+1)^2 du \rightarrow \int u^5(u^4 + 2u^2 + 1) du

    = \int u^9 + 2u^7 + u^5 du \rightarrow \left[\frac{u^{10}}{10} + \frac{u^8}{4} + \frac{u^6}{6}\right]

    = \frac{tan^{10}x}{10} + \frac{tan^8x}{4} + \frac{tan^6x}{6}

    What would be a good method to integrate this another way? I've been looking at it and my trig formulas sheet and haven't been able to come up with anything that will work. Any suggestions?
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  2. #2
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    Hello, VitaX!

    Why factor the tangent into so many parts?


    Integrate using two different methods:

    . . \int \tan^5\!x\sec^6\!x\,dx
    Method 1

    We have: . \int\tan^5\!x\,\sec^4\!x\,(\sec^2\!x\,dx)

    . . . . . =\; \int\tan^5\!x\,(\sec^2\!x)^2\,(\sec^2\!x\,dx)

    . . . . . =\;\int\tan^5\!x\,(\tan^2\!x+1)^2\,(\sec^2\!x\,dx)

    . . . . . =\;\int \tan^5\!x\,(\tan^4\!x + 2\tan^2\!x + 1)\,(\sec^2\!x\,dx)

    . . . . . =\;\int(\tan^9\!x + 2\tan^7\!x + \tan^5\!x)\,(\sec^2\!x\,dx)


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

    Substitute: . \int(u^9 + 2u^7 + u^5)\,du \quad\hdots etc.



    Method 2

    We have: . \int\tan^4\!x\,\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\tan^2\!x)^2\,\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\sec^2\!x-1)^2\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\sec^4\!x - 2\sec^2\!x + 1)\,\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\sec^9\!x - 2\sec^7\!x + \sec^5\!x)\,(\sec x\tan x\,dx)


    Let: . u \:=\:\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx

    Substitute: . \int(u^9 - 2u^7 + u^5)\,du \quad\hdots etc.

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  3. #3
    Member VitaX's Avatar
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    I was just taught as a first step to express the the trig values in forms of even exponents, but ya I can see as it was useless here.

    Quote Originally Posted by Soroban View Post
    Hello, VitaX!

    Why factor the tangent into so many parts?


    Method 1

    We have: . \int\tan^5\!x\,\sec^4\!x\,(\sec^2\!x\,dx)

    . . . . . =\; \int\tan^5\!x\,(\sec^2\!x)^2\,(\sec^2\!x\,dx)

    . . . . . =\;\int\tan^5\!x\,(\tan^2\!x+1)^2\,(\sec^2\!x\,dx)

    . . . . . =\;\int \tan^5\!x\,(\tan^4\!x + 2\tan^2\!x + 1)\,(\sec^2\!x\,dx)

    . . . . . =\;\int(\tan^9\!x + 2\tan^7\!x + \tan^5\!x)\,(\sec^2\!x\,dx)


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

    Substitute: . \int(u^9 + 2u^7 + u^5)\,du \quad\hdots etc.



    Method 2

    We have: . \int\tan^4\!x\,\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\tan^2\!x)^2\,\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\sec^2\!x-1)^2\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\sec^4\!x - 2\sec^2\!x + 1)\,\sec^5\!x\,(\sec x\tan x\,dx)

    . . . . . =\;\int(\sec^9\!x - 2\sec^7\!x + \sec^5\!x)\,(\sec x\tan x\,dx)


    Let: . u \:=\:\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx

    Substitute: . \int(u^9 - 2u^7 + u^5)\,du \quad\hdots etc.

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