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Thread: please help determine the following integrals

  1. #1
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    please help determine the following integrals

    $\displaystyle \int (3x-1)(x+2)dx$

    b)$\displaystyle \int cos(7x+2)dx$

    c)$\displaystyle \int \frac {y}{\sqrt {1+2y^2}} dy$

    d)$\displaystyle \int \frac {sin2x}{1+sin^2x}dx$

    e)$\displaystyle \int_{0}^\frac{\pi}{2}{sin2xcosx} dx$

    f)$\displaystyle \int [\ln{4} - \frac {1}{\ln(e^{2x})}]dx$
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  2. #2
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    Quote Originally Posted by joey1 View Post
    $\displaystyle \int (3x-1)(x+2)dx$

    b)$\displaystyle \int cos(7x+2)dx$

    c)$\displaystyle \int \frac {y}{\sqrt {1+2y^2}} dy$

    d)$\displaystyle \int \frac {sin2x}{1+sin^2x}dx$

    e)$\displaystyle \int_{0}^\frac{\pi}{2}{sin2xcosx} dx$

    f)$\displaystyle \int [\ln{4} - \frac {1}{\ln(e^{2x})}]dx$
    (a) : Expand.
    (b) & (c) : Use a substitution.
    (d) & (e) : Rewrite $\displaystyle sin(2x)$ as $\displaystyle 2sin(x)cos(x)$, then use a substitution.
    (f) : Recall that : $\displaystyle \int ( \, f(x) - g(x) \, ) \, dx = \int f(x) \, dx \, - \int g(x) \, dx$ .. and Note that $\displaystyle ln(e^{f(x)}) = f(x)$.

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