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Thread: The Substitution Rule

  1. #1
    Junior Member
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    The Substitution Rule

    Can someone help me with these two problems. Thanks a lot.
    1. integral of cos^3 thelta sin thelta d thelta, u= cos thelta
    this is what I did so far: du/dx= -sin thelta du= -sin thelta dx => -du=sin thelta d thelta ...I dont even know if this is correct. I need to sub. u into the function but dont know how.
    2. integral sec^2(1/x)/x^2 dx , u = 1/x
    I got -ln x du = dx .....I dont know what to do at all
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  2. #2
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    $\displaystyle \int \cos^3{t} \cdot \sin{t} \, dt$

    $\displaystyle u = \cos{t}$

    $\displaystyle du = -\sin{t} \, dt$

    $\displaystyle -\int \cos^3{t} \cdot (-\sin{t}) \, dt$

    sub ...

    $\displaystyle -\int u^3 \, du$

    do the integration.


    $\displaystyle \int \sec^2\left(\frac{1}{x}\right) \cdot \frac{1}{x^2} \, dx$

    $\displaystyle u = \frac{1}{x}$

    $\displaystyle du = -\frac{1}{x^2} \, dx$

    $\displaystyle -\int \sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \, dx$

    sub ...

    $\displaystyle -\int \sec^2{u} \, du$

    do the integration.
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  3. #3
    Junior Member Ziaris's Avatar
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    Here's exactly what you do for the first one so hopefully you can get the method down.

    $\displaystyle \int\cos^3\theta\sin\theta d\theta$. So if you set $\displaystyle u=\cos\theta, \frac{du}{d\theta}=-\sin\theta$. So move over the $\displaystyle d\theta$, $\displaystyle du=-\sin\theta d\theta$. Solve for $\displaystyle d\theta$, $\displaystyle d\theta=-\frac{du}{\sin\theta}$. Substituting this in,

    $\displaystyle \int\cos^3\theta\sin\theta d\theta=\int u^3\frac{-\sin\theta du}{\sin\theta}=-\int u^3 du=-\frac{1}{4}u^4 + C = -\frac{1}{4}\cos^4\theta + C.$

    This follows after you remember what we defined u as.
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