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Math Help - Integration Help

  1. #1
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    Integration Help

    Hello,

    I was wondering if someone could help me solve the integral of the following:

    int(sec^2(sqrt(x))dx).

    Now, I tried letting u = sqrt(x), but that gets me nowhere; I wouldn't be able to integrate by parts, as what would I let dv be? Trying to let u = the whole function does little good...maybe I am missing something (well, obviously I am!).

    Ugh.

    Thanks in advance for your help!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by AfterShock
    Hello,

    I was wondering if someone could help me solve the integral of the following:

    int(sec^2(sqrt(x))dx).

    Now, I tried letting u = sqrt(x), but that gets me nowhere; I wouldn't be able to integrate by parts, as what would I let dv be? Trying to let u = the whole function does little good...maybe I am missing something (well, obviously I am!).

    Ugh.

    Thanks in advance for your help!!
    First change the variable, let: u=\sqrt{x}, then:

    <br />
dx=2u\ du<br />
.

    So the integral becomes:

    <br />
\int (\sec( \sqrt{x})^2 dx=\int \frac{2u}{(\cos(u))^2}du<br />
.

    This can now be integrated by parts to give:

    <br />
\int \frac{2u}{(\cos(u))^2}du= 2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1)<br />

    Now we need the standard integral:

    <br />
\int \frac{1}{(\cos(u))^2}du=\tan(u)<br />

    which when substituted into (1) gives:

    <br />
\int \frac{2u}{(\cos(u))^2}du= 2 u\tan(u)-2\int \tan(u) du<br />
.

    To complete this you need to know that:

    <br />
\int \tan(x) dx = -\ln(\cos(x))<br />

    RonL
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  3. #3
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    Thank you CaptainBlack. I have a few questions regarding it, though, since I truly want to know how to do it.

    I get the first part, but I lose you on this part:

    This can now be integrated by parts to give:

    <br />
2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1)<br />

    I didn't know you could just pull the 2u out like that; and then you have an integral of an integral? I thought you were doing integration by parts (uv - int(vdu)). I believe I understand everything after that line.

    Thanks- it's appreciated!
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  4. #4
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    Integration by parts is often expressed as \int u dv = [uv] - \int v du. We could also express the LHS as \int fg dx where the indefinite integral of g is the function G to write \int fg = fG - \int f'G. If we write G = \int g then we have the formulation \int fg = f \int G - \int f' \int G.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by AfterShock
    Thank you CaptainBlack. I have a few questions regarding it, though, since I truly want to know how to do it.

    I get the first part, but I lose you on this part:

    This can now be integrated by parts to give:

    <br />
2 u\int \frac{1}{(\cos(u))^2}du-2\int \left\{\int}\frac{1}{(\cos(u))^2}du\right\}\ du\ \ \ \dots(1)<br />

    I didn't know you could just pull the 2u out like that; and then you have an integral of an integral? I thought you were doing integration by parts (uv - int(vdu)). I believe I understand everything after that line.

    Thanks- it's appreciated!
    In your notation int(u dv)=uv - int(vdu), though as we are using u as a
    variable let us write this as in(f dg)=f g - int(g df), so here we are putting
    f(u)=2u, and dg(u)=1/cos^2(u), then g(u)=int(1/cos^2(u)), and df(u)=2.

    Then the result follows.

    RonL
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