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Thread: Clueless in front of an integral (it's not a complicated one though)

  1. #1
    MHF Contributor arbolis's Avatar
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    Clueless in front of an integral (it's not a complicated one though)

    The problem states to use the substitution $\displaystyle t=\tan \left( \frac{x}{2} \right)$ or equivalently $\displaystyle x=2 \arctan (t)$ with the integral $\displaystyle \int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\displaystyle \arctan$ function is $\displaystyle \frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $\displaystyle 2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\displaystyle \sqrt{\cos (x)}$...
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    The problem states to use the substitution $\displaystyle t=\tan \left( \frac{x}{2} \right)$ or equivalently $\displaystyle x=2 \arctan (t)$ with the integral $\displaystyle \int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\displaystyle \arctan$ function is $\displaystyle \frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $\displaystyle 2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\displaystyle \sqrt{\cos (x)}$...
    Ok

    Way two

    $\displaystyle \int\frac{dx}{1+\cos(x)}$

    Now we $\displaystyle x=2\arctan(u)$

    So $\displaystyle dx=\frac{2}{u^2+1}$

    Ok so lets see what

    $\displaystyle \cos\left(2\arctan(x)\right)$ is

    We see that

    $\displaystyle \cos(2x)=2\cos^2(x)-1$

    So then $\displaystyle \cos\left(2\arctan(x)\right)=2\cos^2\left(\arctan( x)\right)-1$

    Now you should know that

    $\displaystyle \cos\left(\arctan(x)\right)=\frac{1}{\sqrt{x^2+1}}$

    So

    $\displaystyle \cos^2\left(\arctan(x)\right)=\frac{1}{x^2+1}$

    So

    $\displaystyle \cos\left(2\arctan(x)\right)+1=\bigg[2\cdot\frac{1}{x^2+1}-1\bigg]+1=\frac{2}{x^2+1}$

    So now we see that we are working in u's so

    $\displaystyle \int\frac{2~du}{u^2+1}\cdot\frac{1}{\frac{2}{u^2+1 }}=\int~du=u+C$

    So $\displaystyle x=2\arctan(x)\Rightarrow{u=\tan\left(\frac{x}{2}\r ight)}$

    So

    $\displaystyle \int\frac{dx}{1+\cos(x)}=\tan\left(\frac{x}{2}\rig ht)$
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  3. #3
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    Quote Originally Posted by arbolis View Post
    The problem states to use the substitution $\displaystyle t=\tan \left( \frac{x}{2} \right)$ or equivalently $\displaystyle x=2 \arctan (t)$ with the integral $\displaystyle \int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\displaystyle \arctan$ function is $\displaystyle \frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $\displaystyle 2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\displaystyle \sqrt{\cos (x)}$...
    Do not do that. Just use $\displaystyle 2\cos^2 \tfrac{x}{2} = 1+\cos x$.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    You could also note that

    $\displaystyle \cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos (x)}{2}}$

    Squaring both sides gives us

    $\displaystyle \cos^2\left(\frac{x}{2}\right)=\frac{1+\cos(x)}{2}$

    So $\displaystyle 2\cos^2\left(\frac{x}{2}\right)=1+\cos(x)$

    EDIT: Looks like TPH beat me to the punch
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by arbolis View Post
    but I don't understand why we bother with a coefficient of $\displaystyle 2$ here...
    Because with $\displaystyle t=\tan \frac{x}{2}$ one has $\displaystyle \cos x=\frac{1-t^2}{1+t^2}$. (trig. identity)

    $\displaystyle x=2\arctan t \implies \mathrm{d}x=\frac{2}{1+t^2}\,\mathrm{d}t$

    $\displaystyle
    \begin{aligned}
    \int\frac{1}{1+\cos x}\,\mathrm{d}x=\int\frac{1}{1+\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\,\mathrm{d}t
    &=\int\frac{2}{1+t^2+\frac{1-t^2}{1+t^2}(1+t^2)}\,\mathrm{d}t\\
    &=\int\frac{2}{2}\,\mathrm{d}t\\
    &=t+C\\
    &=\tan \frac{x}{2}+C\\
    \end{aligned}$
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  6. #6
    MHF Contributor arbolis's Avatar
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    Nice answers, thank you all. I have another integral to do with the same substitution, so I might ask help in another thread . But I hope not.
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  7. #7
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    Multiply & divide by $\displaystyle 1-\cos x.$
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