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Math Help - Integration by parts and recursion formula

  1. #1
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    Integration by parts and recursion formula

    I'm supposed to convert the even powers of tangent to secant using a recursion formula, repeatedly if necessary, and then evaluate.

    Here are to two problems:

    ∫ tan^2(x)sec(x) dx

    ∫ tan^4(x)sec(x) dx


    Here is the recursion formula:

    ∫ sec^n(x) dx = (tan(x)sec^(n-2)(x))/(n-1) - ((n-2)/(n-1))⋅∫ sec^(n-2)(x) dx


    I'm not sure how to go about this, so can someone please help me? Thanks!
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  2. #2
    Member kalyanram's Avatar
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    Re: Integration by parts and recursion formula

    Use the trigonometric relation tan^2x = sec^2x - 1 \implies tan^{2k} x = (sec^2 x - 1)^k now use the binomial expansion. Apply this to the above integrals.

    Second integral becomes
    \int tan^4 x sec x dx = \int (sec^2 x -1)^2 sec x dx = \int sec^5 x dx -2\int sec^3 x dx + \int sec x dx Now use the recursion relationship on each integral.

    Apply the same principle to the first integral as well.
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  3. #3
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    Re: Integration by parts and recursion formula

    Hello, Preston019!

    There is a typo in the Recursion Formua.
    There should be a plus (+) between the two expressions.


    I'm supposed to convert the even powers of tangent to secant using a recursion formula,
    repeatedly if necessary, and then evaluate.

    \text{Recursion Formula: }\:\int\sec^n(x)\,dx \;=\;\frac{1}{n-1}\sec^{n-2}(x)\tan(x) \;{\color{red}+}\; \frac{n-2}{n-1}\int\sec^{n-2}(x)\,dx

    \text{Problem: }\:I \;=\;\int\tan^2(x)\sec(x)\,dx

    \text{We have: }\:I \;=\;\int\left[\sec^2(x) - 1\right]\sec(x)\,dx
    . . . . . . . . . . =\;\int\left[\sec^3(x) - \sec(x)\right]\,dx
    . . . . . . . . . . . =\; \underbrace{\int\sec^3(x)\,dx}_{[1]} - \underbrace{\int\sec(x)\,dx}_{[2]}

    Integral [2] has a standard formula: . \ln|\sec(x) + \tan(x )| + C


    We will apply the Recursion Formula to integral [1].

    Substitute n = 3:

    \int\sec^3(x)\,dx \;=\;\tfrac{1}{2}\sec(x)\tan(x) + \tfrac{1}{2}\int\sec(x)\,dx

    . . . . . . . . . =\;\tfrac{1}{2}\sec(x)\tan(x) + \tfrac{1}{2}\ln|\sec(x) + \tan(x)|


    \text{Then: }\:I \:=\:\tfrac{1}{2}\sec(x)\tan(x) + \tfrac{1}{2}\ln|\sec(x) + \tan(x)| - \ln|\sec(x) + \tan(x)|+C

    . . . . . . =\;\tfrac{1}{2}\sec(x)\tan(x) - \tfrac{1}{2}\ln|\sec(x) + \tan(x)| + C
    Thanks from Preston019
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