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Math Help - Simple Integral question w/ parts

  1. #1
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    Smile Simple Integral question w/ parts

    Okay, I have a question. Is \int arctan{5y} equal to \frac{1}{1+5y^2}\times10y?
    I am working on this problem,
    \int x^2arctan{2x}dx, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.
    I made u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}
    Am I heading in the right direction? Thanks in advance.

    -Matt
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by matt3D View Post
    Okay, I have a question. Is \int arctan{5y} equal to \frac{1}{1+5y^2}\times10y?
    No, we have \frac{\mathrm{d}}{\mathrm{d}y}\left[\arctan (5y)\right]=\frac{5}{1+25y^2} or \arctan (5y) = \int_0^{5y}\frac{1}{1+t^2}\,\mathrm{d}t.
    I am working on this problem,
    \int x^2arctan{2x}dx, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.
    Write [tex]\tan^{-1}[/tex] or [tex]\arctan[/tex] .
    I made u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}
    Am I heading in the right direction?
    I don't think so since you don't know an antiderivative of \arctan. One possibility is to see that if you let v=\arctan (2x) and u'=x^2 then v'=\frac{2}{1+4x^2} and u=\frac{x^3}{3} so uv' is a rational fraction : we don't have to integrate \arctan anymore.

    Does it help ?
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  3. #3
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    Quote Originally Posted by matt3D View Post
    Okay, I have a question. Is \int arctan{5y} equal to \frac{1}{1+5y^2}\times10y?
    I am working on this problem,
    \int x^2arctan{2x}dx, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.
    I made u=2y,\ du=2dy,\ dv=arctan{5y},\ and\ v=\frac{10y}{a+5y^2}
    Am I heading in the right direction? Thanks in advance.

    -Matt
    According to my book of integrals, an integral of \arctan x...

    \int \arctan x\, dx = x\arctan x - \frac{1}{2} \log (1+x^2).

    Does that help?
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  4. #4
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    Oops, I mixed my problem up. This is what I am currently trying to figure out: \int2ytan^{-1}{5y}dy
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  5. #5
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    Okay so, u=\ 2y,\ du=\ 2dy,\ dv=\ tan^{-1}{5y},\ and\ v=5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^4)}+c correct? This is looking more difficult then I thought! So now: 2y\times 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^4)}-\frac {1}{2}\int 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^4)}dy I put the \frac {1}{2} in front of the integral to get rid of the 2 from du which is 2dy...
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matt3D View Post
    Okay so, u=\ 2y,\ du=\ 2dy,\ dv=\ tan^{-1}{5y},\ and\ v=5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^{\color{red}4})}+c correct? This is looking more difficult then I thought! So now: 2y\times 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^{\color{red}4})}-\frac {1}{2}\int 5ytan^{-1}{5y}-\frac {1}{2}\ln{(1+25y^{\color{red}4})}dy I put the \frac {1}{2} in front of the integral to get rid of the 2 from du which is 2dy...
    The terms in red should be 2, not 4.

    Now, from here, do the following:

    We found that \int 5y\tan^{-1}(5y)\,dy= 10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}-\frac {1}{2}{\color{red}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}

    The integral reappears on the right side! So get it on the left side!

    \implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\  ln{(1+25y^2)}\,dy}

    Before we get our answer, try to evaluate the integral I highlighted in red.

    Hint : apply integration by parts! Let u=\ln(1+25y^2).....
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  7. #7
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    \int 2y \tan^{-1} (5y) \ dy

    First, let's make the sub: \begin{array}{lll} s = 5y &  \Rightarrow & ds = 5 dy \iff dy = \displaystyle \frac{ds}{5} \\ y = \displaystyle \frac{s}{5} \end{array}

    and we get: \int 2 \left(\frac{s}{5}\right) \tan^{-1} s \frac{ds}{5} \ \ = \ \ \frac{1}{25} \int 2s \tan^{-1}s \ ds

    Now you can use integration by parts: \begin{array}{ll} u = \tan^{-1} s & dv = 2s ds \\ du = \displaystyle \frac{1}{1+s^2} \ ds & v = s^2 \end{array}
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  8. #8
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    Quote Originally Posted by Chop Suey View Post
    Bad news: tan isn't a variable.

    The 5y\tan^{-1}{5y} is an antiderivative of \tan^{-1}{5y}. Try differentiating and see for yourself.

    You have to be wise when you choose what u and dv should be. You're integrating \tan^{-1}{5y}, but you don't know its integral, so how could you think of it as dv?!
    Yea, I see what you mean!
    Quote Originally Posted by Chris L T521 View Post
    The terms in red should be 2, not 4.

    Now, from here, do the following:

    We found that \int 5y\tan^{-1}(5y)\,dy= 10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}-\frac {1}{2}{\color{red}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}

    The integral reappears on the right side! So get it on the left side!

    \implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\  ln{(1+25y^2)}\,dy}

    Before we get our answer, try to evaluate the integral I highlighted in red.

    Hint : apply integration by parts! Let u=\ln(1+25y^2).....
    How does it become 3/2 on the left side? I know its because you put the integral to the left though like you said, but how exactly?
    Quote Originally Posted by o_O View Post
    \int 2y \tan^{-1} (5y) \ dy

    First, let's make the sub: \begin{array}{lll} s = 5y &  \Rightarrow & ds = 5 dy \iff dy = \displaystyle \frac{ds}{5} \\ y = \displaystyle \frac{s}{5} \end{array}

    and we get: \int 2 \left(\frac{s}{5}\right) \tan^{-1} s \frac{ds}{5} \ \ = \ \ \frac{1}{25} \int 2s \tan^{-1}s \ ds

    Now you can use integration by parts: \begin{array}{ll} u = \tan^{-1} s & dv = 2s ds \\ du = \displaystyle \frac{1}{1+s^2} \ ds & v = s^2 \end{array}
    Okay, I never thought of it like that, I'm going to try that now.
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  9. #9
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    The terms in red should be 2, not 4.

    Now, from here, do the following:

    We found that \int 5y\tan^{-1}(5y)\,dy= 10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}{\color{red}-\tfrac{1}{2}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}

    The integral reappears on the right side! So get it on the left side!

    \implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\  ln{(1+25y^2)}\,dy}

    Before we get our answer, try to evaluate the integral I highlighted in red.

    Hint : apply integration by parts! Let u=\ln(1+25y^2).....
    Quote Originally Posted by matt3D View Post

    How does it become 3/2 on the left side? I know its because you put the integral to the left though like you said, but how exactly?
    We see that we have the term -\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy on the right side of the equation and \int 5y\tan^{-1}(5y)\,dy on the left side.

    Add \tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy to both sides. The integral will disappear on the right and on the left side of the equation we get \int 5y\tan^{-1}(5y)\,dy+\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy=\tfrac{3}{2}\int (5y\tan^{-1}(5y))\,dy

    I hope this makes sense.

    --Chris
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  10. #10
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    Quote Originally Posted by Chris L T521 View Post
    We see that we have the term -\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy on the right side of the equation and \int 5y\tan^{-1}(5y)\,dy on the left side.

    Add \tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy to both sides. The integral will disappear on the right and on the left side of the equation we get \int 5y\tan^{-1}(5y)\,dy+\tfrac{1}{2}\int (5y\tan^{-1}(5y))\,dy=\tfrac{3}{2}\int (5y\tan^{-1}(5y))\,dy

    I hope this makes sense.

    --Chris
    Ah, I see. Thank you. I should have just remembered the 2/2 on the left to add to 1/2
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  11. #11
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    Quote Originally Posted by o_O View Post
    \int 2y \tan^{-1} (5y) \ dy

    First, let's make the sub: \begin{array}{lll} s = 5y &  \Rightarrow & ds = 5 dy \iff dy = \displaystyle \frac{ds}{5} \\ y = \displaystyle \frac{s}{5} \end{array}

    and we get: \int 2 \left(\frac{s}{5}\right) \tan^{-1} s \frac{ds}{5} \ \ = \ \ \frac{1}{25} \int 2s \tan^{-1}s \ ds

    Now you can use integration by parts: \begin{array}{ll} u = \tan^{-1} s & dv = 2s ds \\ du = \displaystyle \frac{1}{1+s^2} \ ds & v = s^2 \end{array}
    So, I get s^2(tan^{-1}s)-\int \frac {s^2}{1+s^2}ds but I need some help integrating the integral, it looks very similar to tan^{-1}{s} except with the s^2 on the numerator.
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  12. #12
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matt3D View Post
    So, I get s^2(tan^{-1}s)-\int \frac {s^2}{1+s^2}ds but I need some help integrating the integral, it looks very similar to tan^{-1}{s} except with the s^2 on the numerator.
    One way to deal with the integral would be to apply long division to the integrand:

    I leave it for you to verify that \frac{s^2}{1+s^2}=1-\frac{1}{1+s^2}

    Now integrate \int\left(1-\frac{1}{1+s^2}\right)\,ds. It should be a piece of cake now.

    I hope this makes sense!

    --Chris
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  13. #13
    Super Member 11rdc11's Avatar
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    Try \int{ds} - \int\frac{ds}{s^2+1} = \int\frac{s^2}{1+s^2}

    oops I was too slow
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  14. #14
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    Alright, so I finally get: \frac {25y^2(tan^{-1}{5y})+tan^{-1}{5y}-5y}{25} Now I'm a little confused about the division though, 11rdc11, could you explain your last post a little more?
    Thank you guys for all of your help! I really do appreciate it.
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  15. #15
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matt3D View Post
    Alright, so I finally get: \frac {25y^2(tan^{-1}{5y})+tan^{-1}{5y}-5y}{25}+{\color{red}C} Now I'm a little confused about the division though, 11rdc11, could you explain your last post a little more?
    Thank you guys for all of your help! I really do appreciate it.


    Where are you stuck in the division process?

    --Chris
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