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Thread: U Substitution

  1. #1
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    U Substitution

    i'm confused on which item should be my u and which should be my du, i've tried a few values on all problems but I feel my answer is wrong and that I'm overlooking something, if you could please help I would be eternally grateful!


    Use U substitution to solve the following:
    a)anti derivative of cos(x)/sin^2(x) dx top valuei/2 bottom value: pi/4
    b) (4(tan^-1(x))^2)/(1+x^2) dx top value: 1 bottom value: -1
    c) 5x(3x^2 - 12)^9 dx top value: 4 bottom value: 0
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  2. #2
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    Quote Originally Posted by Lupus7874 View Post
    i'm confused on which item should be my u and which should be my du, i've tried a few values on all problems but I feel my answer is wrong and that I'm overlooking something, if you could please help I would be eternally grateful!


    Use U substitution to solve the following:
    a)anti derivative of cos(x)/sin^2(x) dx top valuei/2 bottom value: pi/4
    b) (4(tan^-1(x))^2)/(1+x^2) dx top value: 1 bottom value: -1
    c) 5x(3x^2 - 12)^9 dx top value: 4 bottom value: 0
    a) Substitute $\displaystyle u = \sin x$.
    b) Substitute $\displaystyle u = \tan^{-1} x$.
    c) Substitute $\displaystyle u = 3x^2 - 12$.

    If you need more help, please post all your work and state exactly where you're stuck.
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    Ok for example a) I solve to the point of du*u^2 in said integral, but I'm just lost as to what to do when calculating the antiderivative of a product, surely I can't just apply the power rule in reverse can I?
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    Quote Originally Posted by Lupus7874 View Post
    Ok for example a) I solve to the point of du*u^2 in said integral, but I'm just lost as to what to do when calculating the antiderivative of a product, surely I can't just apply the power rule in reverse can I?
    $\displaystyle \int_{\frac{\pi}{4}}^{\pi}\frac{cos(x)}{sin^2(x)}d x$

    If $\displaystyle u=sin(x)$ then $\displaystyle du=cos(x)dx$ so $\displaystyle dx=\frac{du}{cos(x)}$. Substitute these values into the original integral to obtain $\displaystyle \int_{u_1}^{u_2}\frac{1}{u^2}du$. The limits of the integral are determined by the values of $\displaystyle u$ at the corresponding values of $\displaystyle x$.

    This isn't a product. This can be evaluated using the basic formula:

    $\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$
    -----------------------------------------------------------------------------------------------------------------
    Secondly, there isn't a product rule for integration. There's only a sum and diffference rule. For product and quotients you have to deal with on a case-by-case basis. Sometimes you need to use either the substitution rule, integration by parts, partial fractions, etc. For example:

    $\displaystyle \int sec^3(x)tan(x)dx$

    We know that $\displaystyle \frac{d}{dx}sec(x)=sec(x)tan(x)$, so if we use $\displaystyle u=sec(x)$ then $\displaystyle du=sec(x)tan(x)dx$. This means that we can substitute $\displaystyle dx=\frac{du}{sec(x)tan(x)}$ into the original equation, canceling out the $\displaystyle tan(x)$ and leaving only $\displaystyle sec^2(x)$ which is equal to $\displaystyle u^2$.

    Now we have $\displaystyle \int u^2du=\frac{1}{3}u^3+C=\frac{1}{3}sec^3(x)+C$

    Therefore:

    $\displaystyle \int sec^3(x)tan(x)dx=\frac{1}{3}sec^3(x)+C$
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