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    Hey all. For part c of this question I first expanded the brackets to give

    \int_{\frac{1}{3}\pi}^{\frac{5}{  3}\pi}(1-4cost+4cos^2t)dt

    Then, because I've memorised the integral of cos^2x to be

    \frac{x+sinxcox}{2}, I simply integrated to give

    \left [ t-4sint+2(t+sintcost) \right ]_{\frac{1}{3} \pi}^{\frac{5}{3} \pi}

    which further simplifies into

    \left [ 3t+sin2t-4sint \right ]_{\frac{1}{3} \pi}^{\frac{5}{3} \pi}

    after plugging in the values of t, it works out to

    4\pi +3\sqrt3

    which is the correct answer, but the mark scheme says that there must be a correct use of a double angle formula, which I didn't use.
    Does this mean that you can't just memorize integrals such as cos^2x or sin^2x even if they lead you to the correct answer?

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    Attachment 539723539725
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    (Original post by The_Big_E)
    Hey all. For part c of this question I first expanded the brackets to give

    \int_{\frac{1}{3}\pi}^{\frac{5}{  3}\pi}(1-4cost+4cos^2t)dt

    Then, because I've memorised the integral of cos^2x to be

    \frac{x+sinxcox}{2}, I simply integrated to give

    \left [ t-4sint+2(t+sintcost) \right ]_{\frac{1}{3} \pi}^{\frac{5}{3} \pi}

    which further simplifies into

    \left [ 3t+sin2t-4sint \right ]_{\frac{1}{3} \pi}^{\frac{5}{3} \pi}

    after plugging in the values of t, it works out to

    4\pi +3\sqrt3

    which is the correct answer, but the mark scheme says that there must be a correct use of a double angle formula, which I didn't use.
    Does this mean that you can't just memorize integrals such as cos^2x or sin^2x even if they lead you to the correct answer?

    Attachment 539723539725
    Attachment 539723539725
    Technically speaking, you still used the double angle formula correctly.
 
 
 
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