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Math Help - trigonometric integration

  1. #1
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    trigonometric integration

    I have this example

    \int \frac{1}{\sqrt{x^2+25}}dx

    let x=tan(u), then dx=5sec^2(u)du

    \Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du

    \Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C

    now here is where the problem for me lies.

    tan(u)=\frac{x}{5} I understand this but something I am not seeing is how sec(u)=\frac{\sqrt{x^2+5}}{5}

    for the answer

    ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C
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  2. #2
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    Re: trigonometric integration

    Quote Originally Posted by Jonroberts74 View Post
    I have this example
    \int \frac{1}{\sqrt{x^2+25}}dx
    let x=tan(u), then dx=5sec^2(u)du
    \Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du
    \Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C
    now here is where the problem for me lies.
    tan(u)=\frac{x}{5} I understand this but something I am not seeing is how sec(u)=\frac{\sqrt{x^2+5}}{5}
    for the answer ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C
    Draw a right triangle in which there is an acute angle u and \tan(u)=\frac{x}{5}

    Now what is the sec(u)~?
    Thanks from Jonroberts74
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  3. #3
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    Re: trigonometric integration

    Quote Originally Posted by Jonroberts74 View Post
    I have this example

    \int \frac{1}{\sqrt{x^2+25}}dx

    let x=tan(u), then dx=5sec^2(u)du

    \Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du

    \Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C

    now here is where the problem for me lies.

    tan(u)=\frac{x}{5} I understand this but something I am not seeing is how sec(u)=\frac{\sqrt{x^2+5}}{5}

    for the answer

    ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C


    \tan^2(x) + 1 = \sec^2(x)

    \tan(u)=\frac{x}{5}

    \tan^2(u)+1=\sec^2(u)=\frac{x^2}{25}+1

    \sec(u)=\sqrt{\frac{x^2}{25}+1}=\frac{\sqrt{x^2+25  }}{5}

    I think if you check you'll see that it's 25 not 5. My table shows

    \int\frac{1}{\sqrt{x^2+a^2}}dx=\log\left|x+\sqrt{x  ^2+a^2}\right|
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  4. #4
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    Re: trigonometric integration

    Quote Originally Posted by Jonroberts74 View Post
    I have this example

    \int \frac{1}{\sqrt{x^2+25}}dx

    let x=tan(u), then dx=5sec^2(u)du

    \Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du

    \Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C

    now here is where the problem for me lies.

    tan(u)=\frac{x}{5} I understand this but something I am not seeing is how sec(u)=\frac{\sqrt{x^2+5}}{5}

    for the answer

    ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C
    Actually, you let \displaystyle \begin{align*} x = 5\tan{(u)} \end{align*}. Also if \displaystyle \begin{align*} \tan{(u)} = \frac{x}{5} \end{align*} then \displaystyle \begin{align*} \sec{(u)} = \frac{\sqrt{x^2 + 25}}{5} \end{align*}.

    As for your other question, by Pythagoras we have

    \displaystyle \begin{align*} \sin^2{(u)} + \cos^2{(u)} &\equiv 1 \\ \frac{\sin^2{(u)} + \cos^2{(u)}}{\cos^2{(u)}} &\equiv \frac{1}{\cos^2{(u)}} \\ \frac{\sin^2{(u)}}{\cos^2{(u)}} + \frac{\cos^2{(u)}}{\cos^2{(u)}} &\equiv \frac{1}{\cos^2{(u)}} \\ \tan^2{(u)} + 1 &= \sec^2{(u)} \end{align*}

    From here you should be able to see why \displaystyle \begin{align*} \tan{(u)} = \frac{x}{5} \implies \sec{(u)} = \frac{\sqrt{x^2 + 25}}{5} \end{align*}.
    Thanks from Jonroberts74
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  5. #5
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    Re: trigonometric integration

    oops, type. meant x=5tan(u)

    I was over-thinking that one. thanks
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