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Thread: trigonometric integration

  1. #1
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    trigonometric integration

    I have this example

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

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

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

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

    now here is where the problem for me lies.

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

    for the answer

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

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

    Now what is the $\displaystyle sec(u)~?$
    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

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

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

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

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

    now here is where the problem for me lies.

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

    for the answer

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


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

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

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

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

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

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

    Quote Originally Posted by Jonroberts74 View Post
    I have this example

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

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

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

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

    now here is where the problem for me lies.

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

    for the answer

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

    As for your other question, by Pythagoras we have

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

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

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

    I was over-thinking that one. thanks
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