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    How would I find the integral of √5cos(2x - λ) - 1 ?
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    (Original post by znx)
    How would I find the integral of √5cos(2x - λ) - 1 ?
    \displaystyle \int \sqrt{5}\cos (2x-\lambda) -1 .dx = \sqrt{5} \int \cos (2x-\lambda) .dx - \int 1.dx

    For the first integral, you can do it without much thinking. Though if you really must, use the sub u=2x-\lambda
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    (Original post by RDKGames)
    \displaystyle \int \sqrt{5}\cos (2x-\lambda) -1 .dx = \sqrt{5} \int \cos (2x-\lambda) .dx - \int 1.dx

    For the first integral, you can do it without much thinking. Though if you really must, use the sub u=2x-\lambda
    I got (√5/2)sin(2x - λ) - x, so Im supposed to show that when I evaluate it with upper bound λ and lower bound 0, the area is 2 - λ. I cant seem to get the right answer? I thought maybe I had the wrong integral but im not sure now.
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    (Original post by znx)
    I got (√5/2)sin(2x - λ) - x, so Im supposed to show that when I evaluate it with upper bound λ and lower bound 0, the area is 2 - λ. I cant seem to get the right answer? I thought maybe I had the wrong integral but im not sure now.
    Yeah you're not getting that answer from \displaystyle \int_0^{\lambda} \sqrt{5}\cos (2x-\lambda) -1 .dx = \sqrt{5}\sin \lambda - \lambda
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    (Original post by RDKGames)
    Yeah you're not getting that answer from \displaystyle \int_0^{\lambda} \sqrt{5}\cos (2x-\lambda) -1 .dx = \sqrt{5}\sin \lambda - \lambda
    Oh right, the original equation is f(x) = 2(2cosx - sinx)sinx.
    If I integrated that would It come out as 2 - λ?
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    (Original post by znx)
    Oh right, the original equation is f(x) = 2(2cosx - sinx)sinx.
    If I integrated that would It come out as 2 - λ?
    Nope

    What's the FULL question that you're attempting to do here?
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    QUOTE=RDKGames;76291558]Nope

    What's the FULL question that you're attempting to do here?[/QUOTE]

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    Its part iv)
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    (Original post by znx)
    Its part iv)
    You need to use the fact that \tan \lambda = 2 and deduce the exact value of \sin \lambda before substituting it into \sqrt{5} \sin \lambda - \lambda

    EDIT: You actually find it in part (i) anyway so use it.
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    (Original post by RDKGames)
    You need to use the fact that \tan \lambda = 2 and deduce the exact value of \sin \lambda before substituting it into \sqrt{5} \sin \lambda - \lambda

    EDIT: You actually find it in part (i) anyway so use it.
    With the integral what happened to the 1/2 in √5/2?
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    (Original post by znx)
    With the integral what happened to the 1/2 in √5/2?
    Do the integral. You should end up with -\sin(-\lambda) from the lower bound. Use the fact that \sin(-x) \equiv - \sin x so then you just have \sin \lambda from the lower bound. The half shall then cancel accordingly.
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    (Original post by RDKGames)
    Do the integral. You should end up with -\sin(-\lambda) from the lower bound. Use the fact that \sin(-x) \equiv - \sin x so then you just have \sin \lambda from the lower bound. The half shall then cancel accordingly.
    Thank you for taking the time to help, I really appreciate it mate.
 
 
 
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