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Math Help - four basic/easy integrals

  1. #1
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    four basic/easy integrals

    1 \int (x^4+3)^4x^3dx

    u = x^4+3

    du = 4x^3dx

    \frac{1}{4}du = x^3dx

    \frac{1}{4}\int (u)^4 du

    \frac{1}{4}\int \frac{1}{5}(u)^5 + c

    final answer: \frac{1}{20}(x^4+3)^5 + C


    THE NEXT THREE I AM STRUGGLING WITH... HELP IS APPRECIATED!

    2. \int (x^2-2)^2dx

    3. \int sin^5(2x)cos(2x)dx

    4. \int \frac{9x^6+5x^4+7}{x^3}dx
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  2. #2
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    Quote Originally Posted by golfman44 View Post
    1 \int (x^4+3)^4x^3dx

    u = x^4+3

    du = 4x^3dx

    \frac{1}{4}du = x^3dx

    \frac{1}{4}\int (u)^4 du

    \frac{1}{4}\int \frac{1}{5}(u)^5 + c

    final answer: \frac{1}{20}(x^4+3)^5 + C


    THE NEXT THREE I AM STRUGGLING WITH... HELP IS APPRECIATED!

    2. \int (x^2-2)^2dx

    3. \int sin^5(2x)cos(2x)dx

    4. \int \frac{9x^6+5x^4+7}{x^3}dx

    2. Expand.

    \int{(x^2 - 2)^2\,dx} = \int{x^4 - 4x^2 + 4\,dx}

     = \frac{1}{5}x^5 - \frac{4}{3}x^3 + 4x + C.


    3. Use a u substitution.

    \int{\sin^5{(2x)}\cos{(2x)}\,dx} = \frac{1}{2}\int{\sin^5{(2x)}\cdot 2\cos{(2x)}\,dx}.

    Let u = \sin{(2x)} so that \frac{du}{dx} = 2\cos{(2x)}.

    So the integral becomes

    \int{u^5\,\frac{du}{dx}\,dx} = \int{u^5\,du}

     = \frac{1}{6}u^6 + C

     = \frac{1}{6}\sin^6{(2x)} + C.


    4. Divide everything by the denominator.

    \int{\frac{9x^6 + 5x^4 + 7}{x^3}\,dx} = \int{9x^3 + 5x + 7x^{-3}\,dx}

     = \frac{9}{4}x^4 + \frac{5}{2}x^2 - \frac{7}{2}x^{-2} + C

     = \frac{9x^4}{4} + \frac{5x^2}{2} - \frac{7}{2x^2} + C

     = \frac{9x^6 + 10x^4 - 14}{4x^2} + C.
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