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Math Help - U-Substitution Integrals

  1. #1
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    U-Substitution Integrals

    Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

    (x+1)^3/(x^2)dx

    (x+1)/(x^2) dx

    1/(1+4(x^2)) dx

    x(x-10)^10 dx

    (cosx)/(1+(sinx)^2) dx

    (1+tanx)^3/(cosx)^2

    Thanks
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    Quote Originally Posted by ali2141 View Post
    Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

    (x+1)^3/(x^2)dx

    (x+1)/(x^2) dx

    1/(1+4(x^2)) dx

    x(x-10)^10 dx

    (cosx)/(1+(sinx)^2) dx

    (1+tanx)^3/(cosx)^2

    Thanks
    I have no idea what cancellation you are referring to.
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    Quote Originally Posted by ali2141 View Post

    (x+1)/(x^2) dx
    You may not need a substitution for this one.

     <br />
\int \frac{x+1}{x^2} ~dx = \int \frac{x}{x^2} +\frac{1}{x^2}~dx = \int \frac{1}{x} +\frac{1}{x^2}~dx<br />
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    Quote Originally Posted by ali2141 View Post
    Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

    (x+1)^3/(x^2)dx

    (x+1)/(x^2) dx

    1/(1+4(x^2)) dx

    x(x-10)^10 dx

    (cosx)/(1+(sinx)^2) dx

    (1+tanx)^3/(cosx)^2

    Thanks
    It's not clear what you did here. If you want to integrate \int\frac{(x+1)^3}{x^2}dx, you can just expand the numerator \int\frac{x^3+3x^2+3x+1}{x^2}dx and then split it up into a sum. Then you should be able to apply the general formula for polynomial integrands.
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  5. #5
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    I was trying to use u-substitution.

    Wait, so you mean do: x^3/x^2 + 3x^2/x^2 + 3x/x^2 + 1/x^2 ?
    What would be the answer to that then?
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    Quote Originally Posted by ali2141 View Post
    I was trying to use u-substitution.

    Wait, so you mean do: x^3/x^2 + 3x^2/x^2 + 3x/x^2 + 1/x^2 ?
    What would be the answer to that then?
    You don't see how to integrate that? If you're studying the substitution rule, I assume you must be familiar with the basic methods of integral calculus. Just use the sum rule, and integrate each term using the formula \int ax^ndx=\frac{a}{n+1}x^{n+1}+C
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    No, I get that if you have x^3, the antiderivative would be x^4/4
    But I don't know how that would apply to something like x^3/x^2
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    Quote Originally Posted by ali2141 View Post
    But I don't know how that would apply to something like x^3/x^2

     <br />
\frac{x^3}{x^2} = \frac{x\times x \times x}{x \times x} = x<br />
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  9. #9
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    oh wow.
    okay, i feel like an idiot now.
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    Quote Originally Posted by adkinsjr View Post
    You don't see how to integrate that? If you're studying the substitution rule, I assume you must be familiar with the basic methods of integral calculus. Just use the sum rule, and integrate each term using the formula \int ax^ndx=\frac{a}{n+1}x^{n+1}+C
    n dont equal -1
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  11. #11
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    Quote Originally Posted by ali2141 View Post
    Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

    (x+1)^3/(x^2)dx

    (x+1)/(x^2) dx

    1/(1+4(x^2)) dx

    x(x-10)^10 dx

    (cosx)/(1+(sinx)^2) dx

    (1+tanx)^3/(cosx)^2

    Thanks
    The first two have been taken care of.

    for \int{\frac{1}{1 + 4x^2}\,dx} you do not use a u substitution. You use a trigonometric substitution.

    \int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}

     = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}.


    Now use the substitution x = \frac{1}{2}\tan{\theta}.

    Note that \frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}.

    So dx = \frac{1}{2}\sec^2{\theta}\,d\theta.


    Therefore

     = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{  2}\sec^2{\theta}\,d\theta}

     = \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}

     = \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}

     = \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\  theta}\,d\theta}

     = \frac{1}{2}\int{1\,d\theta}

     = \frac{1}{2}\theta + C.


    Now remembering that x = \frac{1}{2}\tan{\theta}

    2x = \tan{\theta}

    \theta = \arctan{2x}.


    So finally our answer is \frac{1}{2}\arctan{2x} + C.
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  12. #12
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    Quote Originally Posted by ali2141 View Post
    (cosx)/(1+(sinx)^2) dx
    substitute u = sinx
    then you will face a famous integral.
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    Quote Originally Posted by ali2141 View Post
    Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

    (x+1)^3/(x^2)dx

    (x+1)/(x^2) dx

    1/(1+4(x^2)) dx

    x(x-10)^10 dx

    (cosx)/(1+(sinx)^2) dx

    (1+tanx)^3/(cosx)^2

    Thanks
    For \int{x(x - 10)^{10}\,dx} you could expand everything, but that could get messy.

    Instead, integration by parts will need to be used.


    Let u = x so that du = 1.

    Let dv = (x - 10)^{10} so that v = \int{(x - 10)^{10}\,dx} = \frac{1}{11}(x - 10)^{11}.


    Therefore \int{x(x - 10)^{10}\,dx} = \frac{1}{11}x(x - 10)^{11} - \int{\frac{1}{11}(x - 10)^{11}\,dx}

     = \frac{1}{11}x(x - 10)^{11} - \frac{1}{132}(x - 10)^{12} + C

     = \frac{1}{11}(x - 10)^{11}\left[x - \frac{1}{12}(x - 10)\right] + C

     = \frac{1}{11}(x - 10)^{11}\left(x - \frac{1}{12}x + \frac{5}{6}\right)

     = \frac{1}{11}(x - 10)^{11}\left(\frac{11}{12}x + \frac{5}{6}\right)

     = \frac{1}{132}(x - 10)^{11}(11x + 10).
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  14. #14
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    Quote Originally Posted by ali2141 View Post
    Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

    (x+1)^3/(x^2)dx

    (x+1)/(x^2) dx

    1/(1+4(x^2)) dx

    x(x-10)^10 dx

    (cosx)/(1+(sinx)^2) dx

    (1+tanx)^3/(cosx)^2

    Thanks
    For \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx}, note that

    \frac{1}{\cos^2{x}} = \sec^2{x}.


    So \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})\sec^2{x}\,dx}

    Now let u = 1 + \tan{x} so that \frac{du}{dx} = \sec^2{x}.

    The integral becomes

    \int{u\,\frac{du}{dx}\,dx}

     = \int{u\,du}

     = \frac{1}{2}u^2 + C

     = \frac{1}{2}(1 + \tan{x})^2 + C.
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  15. #15
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    I am sorry but shouldnot there have been
    <br /> <br />
\int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})^3\sec^2{x}\,dx}<br />
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