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Math Help - 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 t=\tan \left( \frac{x}{2} \right) or equivalently x=2 \arctan (t) with the integral \int \frac{dx}{1+\cos (x)}. I know that the derivative of the \arctan function is \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 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 \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 t=\tan \left( \frac{x}{2} \right) or equivalently x=2 \arctan (t) with the integral \int \frac{dx}{1+\cos (x)}. I know that the derivative of the \arctan function is \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 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 \sqrt{\cos (x)}...
    Ok

    Way two

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

    Now we x=2\arctan(u)

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

    Ok so lets see what

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

    We see that

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

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

    Now you should know that

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

    So

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

    So

    \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

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

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

    So

    \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 t=\tan \left( \frac{x}{2} \right) or equivalently x=2 \arctan (t) with the integral \int \frac{dx}{1+\cos (x)}. I know that the derivative of the \arctan function is \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 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 \sqrt{\cos (x)}...
    Do not do that. Just use 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

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

    Squaring both sides gives us

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

    So 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 2 here...
    Because with t=\tan \frac{x}{2} one has \cos x=\frac{1-t^2}{1+t^2}. (trig. identity)

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

    <br />
\begin{aligned}<br />
\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<br />
&=\int\frac{2}{1+t^2+\frac{1-t^2}{1+t^2}(1+t^2)}\,\mathrm{d}t\\<br />
&=\int\frac{2}{2}\,\mathrm{d}t\\<br />
&=t+C\\<br />
&=\tan \frac{x}{2}+C\\<br />
\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
    Math Engineering Student
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    Multiply & divide by 1-\cos x.
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