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Math Help - Finding the integral

  1. #1
    Member Furyan's Avatar
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    Finding the integral

    Hello

    Just when I thought I was making some progress. I find that I have no idea at all how to approach finding this integral:

    \int\dfrac{\sec^2{x}}{(1 + \tan{x})^3}

    A nudge in the right direction would be very much appreciated.

    Thank you.

    P.S Is this site compromised? Firefox says that it is and I had serious problems trying to post the other day. I recommended the site to a friend of mine but she said she will not use it because Chrome says it's a security risk. That's such a shame, there are so many brilliant people here who can help people who are stuggling with maths. I tried posting this in the appropriate foum, but was not allowed to do so.
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  2. #2
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    Re: Finding the integral

    Quote Originally Posted by Furyan View Post
    Hello

    Just when I thought I was making some progress. I find that I have no idea at all how to approach finding this integral:

    \int\dfrac{\sec^2{x}}{(1 + \tan{x})^3}

    A nudge in the right direction would be very much appreciated.

    Thank you.

    P.S Is this site compromised? Firefox says that it is and I had serious problems trying to post the other day. I recommended the site to a friend of mine but she said she will not use it because Chrome says it's a security risk. That's such a shame, there are so many brilliant people here who can help people who are stuggling with maths. I tried posting this in the appropriate foum, but was not allowed to do so.
    Make the substitution \displaystyle \begin{align*} u = 1 + \tan{x} \implies du = \sec^2{x}\,dx \end{align*}.
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  3. #3
    Member Furyan's Avatar
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    Re: Finding the integral

    Thank you Prove It

    Quote Originally Posted by Prove It View Post
    Make the substitution \displaystyle \begin{align*} u = 1 + \tan{x} \implies du = \sec^2{x}\,dx \end{align*}.
    I have the following:

    \int\sec^2x(1 + tanx)^{-3} dx

    u = 1 + tanx

    \int\sec^2xu^{-3} dx =\int\sec^2xu^{-3}\dfrac{dx}{du} du

    \dfrac{du}{dx} = \sec^2x

    \dfrac{dx}{du} = \dfrac{1}{\sec^2x}

    \int\sec^2xu^{-3}\dfrac{1}{\sec^2x} du = \int u^{-3}du

    = -\dfrac{1}{2}u^{-2} +c

    u = 1 + tanx

    \int\sec^2x(1 + tanx)^{-3} dx = -\dfrac{1}{2}(1+tanx)^{-2} + c

    I stumbled through that a bit.

    Thank you for getting me started.
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  4. #4
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    Re: Finding the integral

    Your answer is correct, though you took the scenic route. You can see that \displaystyle \begin{align*} \sec^2{x}\,dx \end{align*} is already in the integral. Replace it with \displaystyle \begin{align*} du \end{align*}.
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  5. #5
    Member Furyan's Avatar
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    Re: Finding the integral

    Quote Originally Posted by Prove It View Post
    Your answer is correct, though you took the scenic route. You can see that \displaystyle \begin{align*} \sec^2{x}\,dx \end{align*} is already in the integral. Replace it with \displaystyle \begin{align*} du \end{align*}.
    Thank you, I see what you are saying now, but could someone please clarify this:

    I understood that \dfrac{du}{dx} does not mean du divided by dx so why can I rearrange \dfrac{du}{dx} = \sec^2x to get du = \sec^2x dx.

    And if \dfrac{d}{dx} is an operator and \dfrac{du}{dx} means the derivative of u with respect to x. What does du = \sec^2x dx mean?

    Thank you.
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  6. #6
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    Re: Finding the integral

    Quote Originally Posted by Furyan View Post
    Thank you I get that now, but could someone clarify this:

    I understood that \dfrac{du}{dx} does not mean du divided by dx so why can I rearrange \dfrac{du}{dx} = \sec^2x to get du = \sec^2x dx.

    And if \dfrac{d}{dx} is an operator and \dfrac{du}{dx} means the derivative of u with respect to x. What does du = \sec^2x dx mean.
    It's a sloppy shorthand, I'll give you that.

    Really, what happens is that when you replace \displaystyle \begin{align*} \sec^2{x} \end{align*} with \displaystyle \begin{align*} \frac{du}{dx} \end{align*}, in your integral you get

    \displaystyle \begin{align*} \int{\frac{\sec^2{x}}{(1 + \tan{x})^3}\,dx} &= \int{\frac{1}{u^3}\,\frac{du}{dx}\,dx} \\ &= \int{\frac{1}{u^3}\,du} \end{align*}

    which simplifies from the Chain Rule (not from "cancelling the \displaystyle \begin{align*} dx \end{align*}"). However, most people do the sloppy simplification to save time.
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  7. #7
    Member Furyan's Avatar
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    Re: Finding the integral

    Quote Originally Posted by Prove It View Post
    It's a sloppy shorthand, I'll give you that.

    Really, what happens is that when you replace \displaystyle \begin{align*} \sec^2{x} \end{align*} with \displaystyle \begin{align*} \frac{du}{dx} \end{align*}, in your integral you get

    \displaystyle \begin{align*} \int{\frac{\sec^2{x}}{(1 + \tan{x})^3}\,dx} &= \int{\frac{1}{u^3}\,\frac{du}{dx}\,dx} \\ &= \int{\frac{1}{u^3}\,du} \end{align*}

    which simplifies from the Chain Rule (not from "cancelling the \displaystyle \begin{align*} dx \end{align*}"). However, most people do the sloppy simplification to save time.
    Thank you very much
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