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Math Help - reduction formula for integral

  1. #1
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    reduction formula for integral

    using the identity (sec^2 X = 1+tan^2 X) find a reduction formula for the integral  ln=\int tan^n X dx,  where n is greater than or equal to 2

    hence evaluate \int tan^4 X dx from 0 to pie/4 ( cant really type this in the mathematical terms.

    ok to express this in reduction form i made

    tan^2 X = sec^2X-1 (for the identity)

    then substituted it in the expression  ln=\int  sec^2X -1  dx

    did i do it right? and if so how would that help me to evaluate the integral.
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  2. #2
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    Hello, sigma1!

    I'll do the derivation . . .


    Using the identity \sec^2\!x \:= \:1+\tan^2\!x, find a reduction formula for the integral:

    . . \int \tan^n\!x\,dx \;\;\text{ where }n \geq 2

    We have: . \int(\tan x)^n\,dx

    . . . . . =\;\int \tan^2\!x\,(\tan x)^{n-2}\,dx

    . . . . . =\;\int\left(\sec^2\!x-1\right)(\tan x)^{n-2}\,dx

    . . . . . =\;\int \bigg[\sec^2\!x(\tan x)^{n-2} - (\tan x)^{n-2}\bigg]\,dx

    . . . . . =\;\int(\tan x)^{n-2}\sec^2\!x\,dx \;- \int(\tan x)^{n-2}\,dx

    . . . . . =\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx


    Therefore: . \int(\tan x)^n\,dx \;=\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx

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  3. #3
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    thanks for the reply , but i have been trying to follow your steps and i have to admit i got bit confused after the first step. do you mind explaining to me the steps.
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  4. #4
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    Hello again, sigma1!


    We have: . \int(\tan x)^n\,dx

    Factor: . \int \tan^2\!x\,(\tan x)^{n-2}\,dx

    . . . . \int\overbrace{\left(\sec^2\!x-1\right)}(\tan x)^{n-2}\,dx . .
    Trig identity: \tan^2\!x \:=\:\sec^2\!x-1


    Multiply: . \int \bigg[\sec^2\!x(\tan x)^{n-2} - (\tan x)^{n-2}\bigg]\,dx


    Make two integrals: . \int(\tan x)^{n-2}\sec^2\!x\,dx \;- \int(\tan x)^{n-2}\,dx


    . . For the first integral, let: . u\,=\,\tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx

    . . Substitute: . \int u^{n-2}\,du \;=\;\frac{1}{n-1}u^{n-1}

    . . Back-substitute: . \frac{1}{n-1}(\tan x)^{n-1}


    Hence, we have: . =\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx


    Therefore: . \int(\tan x)^n\,dx \;=\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx

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  5. #5
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    THANKS ALOT.

    now that its reduced how would i use that to evaluate \int tan^4x dx using pie/4 as the upper limit and 0 as the lower limit.
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