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Math Help - Integral Question

  1. #1
    Junior Member
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    Sep 2009
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    Integral Question

    2 Part Question.

    1) How do I use the 'fancy' font to describe formulas on this and other websites? It would make things much easier. I assume some kind of Alt-codes?

    2) The actual question: Integral of (secxcos2x)dx/(sinx + secx)

    The thoughts I've had: cos2x = cos^2x - sin^2x.. I could split up the fraction this way, but I don't see how it's helpful? Obviously d/dx sinx = cosx, but I can't seem to get rid of the 2x in cos2x. My algebra isn't too strong, but it's not possible right? secx = 1/cosx, but again, can't use this.

    I'm sure it has something to do with substitution of some sort, integration by parts seems highly doubtful, due to the denominator.

    I'd like tips instead of just the answer, but a tip which sends me firmly in the right direction!

    Thanks so much for your help, this website has been fantastic resource for me.
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  2. #2
    Member
    Joined
    Sep 2009
    Posts
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    It's called LateX, I actually looked at what other people were posting (you can click on it to see how it was done) and just examining it and seeing what comes out. (i'd never used it until I started posting on here btw)

    You use "[tex]" "[ /math]" and type stuff in between.


    LaTeX Equation Editor - SITMO

    that link also allowed me to test different things before I posted. Also talking to people there's a section devoted to helping people with latex, otherwise I just google something like "latex integrals."

    so like if you want a fraction you type in \frac{a}{b} and that'll give you a/b.

    \int gives you an integral

    \int_b^a dx gives you an integral with limits b being the lower limit, a being the upper limit So the "_a" gives you subscript and the "^b" gives you superscript.

    So on and so forth.


    Have you thought about rewriting the denominator?

    \int \frac{sec x cos (2x)}{sin x + sec x}dx

    =\int \frac{\frac{1}{sec x}*cos(2x)}{\frac{cos x sin x + 1}{cos x}}dx

    =\int cosx*\frac{\frac{1}{cos x}*cos(2x)}{cos x sin x + 1}dx

    =\int \frac{cos(2x)}{cosxsinx+1}dx

    Least that's my initial impression. I think the following might be more helpful.

    sin(2\theta)=2sin\theta cos\theta
    Last edited by seld; September 15th 2009 at 09:46 AM. Reason: fixed it so [math] could be shown
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