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Thread: [SOLVED] Integration by parts : existence ??

  1. #1
    Moo
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    [SOLVED] Integration by parts : existence ??

    Hi

    Another question, not mine this time

    --------------------------------------------------
    Say we want to compute $\displaystyle \int \tan(x) ~dx=\int \frac{\sin(x)}{\cos(x)} ~dx$

    Integration by parts :
    $\displaystyle dv=\sin(x) \Rightarrow v=-\cos(x)$
    $\displaystyle u=\frac{1}{\cos(x)} \Rightarrow du=\frac{\sin(x)}{\cos^2(x)}$

    So $\displaystyle \int \frac{\sin(x)}{\cos(x)} ~dx=\left[\frac{-\cos(x)}{\cos(x)}\right]+\int \frac{\sin(x)}{\cos(x)} ~dx=-1+\int \frac{\sin(x)}{\cos(x)} ~dx$

    $\displaystyle \boxed{0=1}$ ???

    What went wrong ?
    (I know we can recognize -ln(cos(x))


    Now, there is the same problem with the integration by parts if we want to compute $\displaystyle \int_0^{\pi/4} \tan(x) ~dx$ (I mean for a definite integral)

    -------------------------------------------------
    Another integral :
    $\displaystyle \int_0^{\pi/2} \frac{1}{1+\cos(x)} ~dx$
    How would you go about this one ? :P
    Apart from the Weierstrass substitution $\displaystyle t=\tan \tfrac x2$ or complex analysis...


    Thaaaaaaaaank you
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Moo View Post
    Hi

    Another question, not mine this time

    --------------------------------------------------
    Say we want to compute $\displaystyle \int \tan(x) ~dx=\int \frac{\sin(x)}{\cos(x)} ~dx$

    Integration by parts :
    $\displaystyle dv=\sin(x) \Rightarrow v=-\cos(x)$
    $\displaystyle u=\frac{1}{\cos(x)} \Rightarrow du=\frac{\sin(x)}{\cos^2(x)}$

    So $\displaystyle \int \frac{\sin(x)}{\cos(x)} ~dx=\left[\frac{-\cos(x)}{\cos(x)}\right]+\int \frac{\sin(x)}{\cos(x)} ~dx=-1+\int \frac{\sin(x)}{\cos(x)} ~dx$

    $\displaystyle \boxed{0=1}$ ???

    What went wrong ?
    Nothing, because 0=1, up to a constant

    The confusion comes for the "hooks" part: $\displaystyle \left[\frac{-\cos}{\cos}\right]=[-1]$. Either this is definite integration, and you substract using the bounds of the integrals, say $\displaystyle (-1)(b)-(-1)(a)=(-1)-(-1)=0$, or this is undefinite integration, and the result gives an antiderivative, which is unique "up to a constant".
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  3. #3
    Moo
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    Quote Originally Posted by Laurent View Post
    Nothing, because 0=1, up to a constant

    The confusion comes for the "hooks" part: $\displaystyle \left[\frac{-\cos}{\cos}\right]=[-1]$. Either this is definite integration, and you substract using the bounds of the integrals, say $\displaystyle (-1)(b)-(-1)(a)=(-1)-(-1)=0$, or this is undefinite integration, and the result gives an antiderivative, which is unique "up to a constant".
    That was... hmmm simple lol ><

    As for the second integral, I've though of the substitution t=pi/2-x
    and then multiply by (1-sin(x))(1-sin(x))
    however, there is a problem for the lower bound :s
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    Hello, Moo!

    Ha! . . . Good one!


    Say we want to compute $\displaystyle \int \tan(x)\,dx \;=\;\int \frac{\sin(x)}{\cos(x)}\,dx$

    Integration by parts: .$\displaystyle \begin{array}{ccccccc}dv\:=\:\sin(x) &\Rightarrow& v\:=\:-\cos(x) \\

    u\:=\:\frac{1}{\cos(x)}& \Rightarrow & du\:=\:\frac{\sin(x)}{\cos^2(x)} \end{array}$


    So: .$\displaystyle \int \frac{\sin(x)}{\cos(x)}\,dx \;=\;\left[\frac{-\cos(x)}{\cos(x)}\right]+\int \frac{\sin(x)}{\cos(x)}\,dx\;=\;-1+\int \frac{\sin(x)}{\cos(x)}\,dx$

    $\displaystyle \boxed{0=1}$ ???

    What went wrong ? (Worried)

    You forgot the "plus C" . . .


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    A classic proof that "1 = 0".


    Integrate: .$\displaystyle \int\sin x\cos x\,dx$


    (a) We have: .$\displaystyle \int\sin x(\cos x\,dx)$

    Let $\displaystyle u \,=\,\sin x \quad\Rightarrow\quad du \,=\,\cos x\,dx$

    Substitute: .$\displaystyle \int u\,du \;=\;\tfrac{1}{2}u^2$

    Back-substitute: .$\displaystyle \tfrac{1}{2}\sin^2\!x$ .[1]



    (b) We have: .$\displaystyle \int\cos x(\sin x\,dx)$

    Let $\displaystyle u \,=\,\cos x\quad\Rightarrow\quad du \,=\,-\sin x\,dx$

    Substitute: .$\displaystyle \int u(-du) \;=\;-\tfrac{1}{2}u^2$

    Back-substitute: .$\displaystyle -\tfrac{1}{2}\cos^2\!x$ .[2]


    Since [1] and [2] are equal, we have: .$\displaystyle \tfrac{1}{2}\sin^2\!x \:=\:-\tfrac{1}{2}\cos^2\!x$

    Therefore: .$\displaystyle \sin^2\!x \:=\:-\cos^2\!x \quad\Rightarrow\quad\sin^2\!x + \cos^2\!x \:=\:0 \quad\Rightarrow\quad 1 \:=\:0$

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  5. #5
    Super Member PaulRS's Avatar
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    $\displaystyle
    \int_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} {{\textstyle{{dx} \over {1 + \cos \left( x \right)}}}} = \int_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} {{\textstyle{{1 - \cos \left( x \right)} \over {\sin ^2 \left( x \right)}}}dx}
    $$\displaystyle
    = - \int_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} {\left( {1 - \cos \left( x \right)} \right) \cdot \left( {\cot \left( x \right)} \right)^\prime dx}
    $

    Now integrate by parts.
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  6. #6
    Moo
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    Quote Originally Posted by Soroban View Post
    You forgot the "plus C" . . .
    Yes, but I think I forgot it because the integral signs were kept... that was stupid

    Quote Originally Posted by PaulRS View Post
    $\displaystyle
    \int_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} {{\textstyle{{dx} \over {1 + \cos \left( x \right)}}}} = \int_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} {{\textstyle{{1 - \cos \left( x \right)} \over {\sin ^2 \left( x \right)}}}dx}
    $$\displaystyle
    = - \int_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 2$}}} {\left( {1 - \cos \left( x \right)} \right) \cdot \left( {\cot \left( x \right)} \right)^\prime dx}
    $

    Now integrate by parts.
    Looks like you've been playing with the latex

    Oh yeah, and $\displaystyle \lim_{x \to 0} (1-\cos(x)) \cot(x)$ must be defined...




    Here is the problem I've come up with ...
    substitute $\displaystyle t=\pi/2-x$
    then the integral is now :
    $\displaystyle \int_0^{\pi/2} \frac{dx}{1+\sin(x)} ~dx=\int_0^{\pi/2} \frac{1}{\cos^2(x)}-\frac{\sin(x)}{\cos^2(x)} ~dx$
    If we separate the difference and compute the two integrals independently, there is a problem because the limit $\displaystyle \frac{1}{\cos(x)}$ when x goes to pi/2 is not defined.
    Is it wrong because since the two terms are not integrable, we can't separate the difference ?
    So I guess it's an integration by parts exactly like in yours that we have to do here ^^'
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    Super Member PaulRS's Avatar
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    Quote Originally Posted by Moo View Post
    Yes, but I think I forgot it because the integral signs were kept... that was stupid


    Looks like you've been playing with the latex

    Oh yeah, and $\displaystyle \lim_{x \to 0} (1-\cos(x)) \cot(x)$ must be defined...
    That limit is obviously 0

    You cannot separate because the integrals are divergent, it'd be like writing $\displaystyle +\infty-\infty=\text{a real number}$.

    Unless you change the integration limits, say to $\displaystyle \pi-\epsilon$ and $\displaystyle \epsilon$ with $\displaystyle 0<\epsilon<1$ and then take $\displaystyle \epsilon\to 0$
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    Moo
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    $\displaystyle \int_0^{\pi/2} \frac{dx}{1+\cos(x)}$

    Substitute $\displaystyle x=2t$ :

    and using the identity $\displaystyle 1+\cos 2t=2 \cos^2(t)$ :
    $\displaystyle \int_0^{\pi/4} \frac{dt}{\cos^2(t)}=[\tan(t)]_0^{\pi/4}=1$

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