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    (1+cos(2x))^2?
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    (Original post by Tynos)
    (1+cos(2x))^2?
    Expand the bracket and integrate term by term.
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    (Original post by Tynos)
    (1+cos(2x))^2?
    Expand brackets and when you have cos^2 (2x) you reverse double angle formula it
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    You would have to expand the bracket. So:
    (1 + cos(2x))^2 = 1^2 + 2(1)(cos(2x)) + (cos(2x))^2 = 1 + 2cos(2x) + cos^2(2x) (by the Binomial Theorem)

    Integrating 1 and 2cos(2x) with respect to x is easy enough, but you will have to use the double angle identity to change cos^2(2x) into a form which you can integrate.

    Now cos(4x) = cos(2(2x)) = cos^2(2x) - sin^2(2x) = 2cos^2(2x) - 1.
    Therefore 2cos^2(2x) = cos(4x) + 1, so cos^2(2x) = 0.5cos(4x) + 0.5

    Therefore, ∫ (1 + cos(2x))^2 dx = ∫ (1 + 2cos(2x) + 0.5cos(4x) + 0.5) dx = ∫ (1.5 + 2cos(2x) + 0.5cos(4x)) dx
    This is equal to 1.5x + sin(2x) + 0.125sin(4x) + c, where c is an arbitrary constant. That's how you integrate that function - I'd advise checking my working to make sure I haven't made any silly mistakes.
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    You go to www.integrals.com
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    Integration is so far the one part of maths which could not be easier, even STEP integrals are fine.,but hey thats my opinion. If you have an odd powered trig function split it up then reverse chain rule:

    i.e. sin^3(x) -> sinx(sin^2(x)) -> sinx(1-cos^2(x)) -> sinx - sinxcos^2(x)

    If it is even powered just use a double angle formulae to reduce the power.
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    (Original post by Jai Sandhu)
    Integration is so far the one part of maths which could not be easier, even STEP integrals are fine.,but hey thats my opinion. If you have an odd powered trig function split it up then reverse chain rule:

    i.e. sin^3(x) -> sinx(sin^2(x)) -> sinx(1-cos^2(x)) -> sinx - sinxcos^2(x)

    If it is even powered just use a double angle formulae to reduce the power.
    i dare u to integrate 1/lnx
    call it easy again
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    (Original post by CancerousProblem)
    i dare u to integrate 1/lnx
    call it easy again



    please
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    Lol the thread title reminds me of this:
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    (Original post by rayquaza17)
    Lol the thread title reminds me of this:
    I loled
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    (Original post by Tiri)
    i dare u to integrate 1/lnx
    call it easy again
    well 1/1 = 1

    so it is ∫ nx dx

    which is nx2/2 + c

    :hat2:
 
 
 
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