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Math Help - Integration

  1. #1
    Member srirahulan's Avatar
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    Arrow Integration

    \int [sec^3x] d(x) I Tried so many times but I cannot find the answer>>>> First I use the law of integration I write this like this \int [secx\frac{dtanx}{d(x)}]d(x) Then I cannot do anything.>>>>
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  2. #2
    MHF Contributor

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    Re: Integration

    What I prefer to do is change everything to sine and cosine. sec(x)= 1/cos(x).
    So \int sec(x)dx= \int\frac{1}{cos^3(x)}dx

    Multiply both numerator and denominator by cos(x) to get
    \int \frac{cos(x)}{cos^4(x)}dx= \int\frac{cos(x)}{(1- sin^2(x))^2} dx

    Now let u= sin(x), du= cos(x)dx and that becomes \int\frac{du}{(1- u^2)^2}

    which can be done by "partial fractions".
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  3. #3
    Junior Member ReneG's Avatar
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    Re: Integration

    Use integration by parts.

    \int \sec^3 x\,dx = \int \sec^2x \sec x \,dx

    let dv = \sec^2x\,dx \Rightarrow v = \tan x
    and u = \sec x \Rghtarrow du = \sec x\tan x \,dx

    \begin{align*}\int \sec^3 x\,dx &= \sec x\tan x - \int \tan^2x\sec x\,dx \\ &= \sec x\tan x- \int(\sec^2x - 1) \sec x\,dx \\ &= \sec x\tan x - \left(\int \sec^3 x \,dx - \int \sec x\,dx\right) \\ 2\int \sec^3x\,dx &= \sec x \tan x + \int \sec x \,dx \\ \\ \int \sec^3 x\,dx &= \frac{1}{2}\left[\sec x \tan x + \int \sec x \,dx \right ]\end{align*}
    Last edited by ReneG; July 5th 2013 at 10:09 AM.
    Thanks from srirahulan
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  4. #4
    Member srirahulan's Avatar
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    Re: Integration

    I have a great thank for both of you.
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