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Thread: Integral, is my result correct?

  1. #1
    MHF Contributor arbolis's Avatar
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    Integral, is my result correct?

    Just need to be sure on this : $\displaystyle \int \frac{dx}{2+\sin x +\cos x}$, making the substitution $\displaystyle x=2 \arctan (t)$ I reached that the integral is worth $\displaystyle \frac{2 \tan \left( \frac{x}{2}\right) }{5}+C$.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Just need to be sure on this : $\displaystyle \int \frac{dx}{2+\sin x +\cos x}$, making the substitution $\displaystyle x=2 \arctan (t)$ I reached that the integral is worth $\displaystyle \frac{2 \tan \left( \frac{x}{2}\right) }{5}+C$.
    It is wrong, and your math intuition should tell you so

    If $\displaystyle \int\frac{dx}{1+\cos(x)}=\tan\left(\frac{x}{2}\rig ht)$

    Then adding a WHOLE NEW FUNCTION to the denominator should not change the answer by a constant.
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  3. #3
    MHF Contributor arbolis's Avatar
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    You're right. I think I know where is my error...
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    You're right. I think I know where is my error...
    Glad to hear it! I knew if you actually thought about it you could answer it yourself ...you don't give yourself enough credit! But if you do have any problems/questions just ask!
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  5. #5
    MHF Contributor arbolis's Avatar
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    You're right. I think I know where is my error... Hmm no...
    Let me show you what I did : Let $\displaystyle x=2\arctan (t)$, $\displaystyle dx=\frac{2}{1+t^2}$.
    So the integral becomes $\displaystyle \int \frac{\frac{2}{1+t^2}}{2+\sin(2\arctan(t))+\cos(2\ arctan(t))}dt$.
    Now let $\displaystyle \theta = \arctan(t)$.
    We know that $\displaystyle \sin(2\theta)=2\cos(\theta)\sin(\theta)$ and that $\displaystyle \cos(2\theta)=2\cos ^2(\theta)-1$. Furthermore, $\displaystyle \cos(\arctan(t))=\frac{1}{\sqrt{1+t^2}}$ and $\displaystyle \sin(\arctan(t))=\frac{t}{\sqrt{1+t^2}}$.
    Therefore we can rewrite the integral as $\displaystyle \int \frac{\frac{2}{1+t^2}}{2+2\left(\frac{1}{\sqrt{1+t ^2}} \right) \left( \frac{t}{\sqrt{1+t^2}} \right) +\frac{2}{1+t^2}-1}dt$, and from it I reached the result I gave in my first post. I thought I might have made an error for $\displaystyle \sin(\arctan(t))$, but I don't think so. Can you help me finding where is my error?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    You're right. I think I know where is my error... Hmm no...
    Let me show you what I did : Let $\displaystyle x=2\arctan (t)$, $\displaystyle dx=\frac{2}{1+t^2}$.
    So the integral becomes $\displaystyle \int \frac{\frac{2}{1+t^2}}{2+\sin(2\arctan(t))+\cos(2\ arctan(t))}dt$.
    Now let $\displaystyle \theta = \arctan(t)$.
    We know that $\displaystyle \sin(2\theta)=2\cos(\theta)\sin(\theta)$ and that $\displaystyle \cos(2\theta)=2\cos ^2(\theta)-1$. Furthermore, $\displaystyle \cos(\arctan(t))=\frac{1}{\sqrt{1+t^2}}$ and $\displaystyle \sin(\arctan(t))=\frac{t}{\sqrt{1+t^2}}$.
    Therefore we can rewrite the integral as $\displaystyle \int \frac{\frac{2}{1+t^2}}{2+2\left(\frac{1}{\sqrt{1+t ^2}} \right) \left( \frac{t}{\sqrt{1+t^2}} \right) +\frac{2}{1+t^2}-1}dt$, and from it I reached the result I gave in my first post. I thought I might have made an error for $\displaystyle \sin(\arctan(t))$, but I don't think so. Can you help me finding where is my error?
    I think your error is in simplifying that integral at the end
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  7. #7
    MHF Contributor arbolis's Avatar
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    I think your error is in simplifying that integral at the end
    I made an error there! Yes...
    Finally I got the answer $\displaystyle \frac{2}{5+\tan ^2(\frac{x}{2})}+C$. I hope this time it's correct.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I made an error there! Yes...
    Finally I got the answer $\displaystyle \frac{2}{5+\tan ^2(\frac{x}{2})}+C$. I hope this time it's correct.
    Check it one more time.
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  9. #9
    MHF Contributor arbolis's Avatar
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    Now I got that it's equal to $\displaystyle \int \frac{2dt}{t^2+2t+3}$. Now I must calculate this.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Now I got that it's equal to $\displaystyle \int \frac{2dt}{t^2+2t+3}$. Now I must calculate this.


    First notice that

    $\displaystyle t^2+2t+3=(t+1)^2+2$

    So we have

    $\displaystyle 2\int\frac{dt}{(t+1)^2+2}$

    Now let $\displaystyle t+1=\sqrt{2}\tan(\theta)$
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  11. #11
    MHF Contributor arbolis's Avatar
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    Before I read
    First notice that



    So we have



    Now let
    I tried by integration by parts and then by substitution but didn't reach anything. Now following your trick, I got that the final result is...................................... $\displaystyle \frac{\tan \left(\frac{x}{2} \right)}{\frac{\tan\left(\frac{x}{2} \right)+1}{\sqrt{2}}}+C$ or $\displaystyle \frac{\sqrt{2}\tan\left(\frac{x}{2} \right)}{\tan\left(\frac{x}{2} \right)+1}+C$. But you know what? I've a strong feeling I made at least an error in all that.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Before I read I tried by integration by parts and then by substitution but didn't reach anything. Now following your trick, I got that the final result is...................................... $\displaystyle \frac{\tan \left(\frac{x}{2} \right)}{\frac{\tan\left(\frac{x}{2} \right)+1}{\sqrt{2}}}+C$ or $\displaystyle \frac{\sqrt{2}\tan\left(\frac{x}{2} \right)}{\tan\left(\frac{x}{2} \right)+1}+C$. But you know what? I've a strong feeling I made at least an error in all that.
    I am going to do the integral for you, so that you can check it when you get stuck ok?


    $\displaystyle 2\int\frac{dt}{(t+1)^2+2}$

    Let $\displaystyle t+1=\sqrt{2}\tan(\theta)$

    $\displaystyle \Rightarrow{dt=\sqrt{2}\sec^2(\theta)}$

    $\displaystyle \therefore\quad2\int\frac{dt}{(t+1)^2+2}\underbrac e{\Rightarrow}_{t+1=\sqrt{2}\tan(\theta)}2\int\fra c{\sqrt{2}\sec^2(\theta)}{2\tan^2(\theta)+2}~d\the ta$

    $\displaystyle =\sqrt{2}\int\frac{\sec^2(\theta)}{\sec^2(\theta)~ }d\theta$

    $\displaystyle =\sqrt{2}\int~d\theta$

    $\displaystyle =\sqrt{2}\theta$

    $\displaystyle \underbrace{\Rightarrow}_{\text{backsub}}2\int\fra c{dt}{(t+1)^2+2}=\sqrt{2}\arctan\left(\frac{t+1}{\ sqrt{2}}\right)$

    $\displaystyle \underbrace{\Rightarrow}_{\text{backsub again}}\int\frac{dx}{2+\sin(x)+\cos(x)}=\sqrt{2}\a rctan\left(\frac{\tan\left(\frac{x}{2}\right)+1}{\ sqrt{2}}\right)+C$
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