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Math Help - Integration by reduction and by parts

  1. #1
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    Thumbs down Integration by reduction and by parts

    I've been chipping away at these two problems for a couple days now - they're part of a much larger assignment that is due today - and I've had no luck whatsoever with them. Every time I think a lightbulb has lit up, I end up with a couple pages of work proving nothing. Here are the two questions, and if anybody has any suggestions they would be greatly appreciated.

    1) I = (integral) x*(arctanx)^2dx

    For this one I'm supposed to solve it in terms of x, but I've tried integrating by parts, using x=([x^2]/2)', and then in the resulting equation substituting tanu for x in the integral.
    It took me from (integral) [(arctanx)x^2]/(1+x^2) dx to (integral)u(tanu)^2du but I think that restricts u to being on +/- pi/2. Either way, I didn't get any useful answer out of it.

    2) I = (integral) [(x^2 + 1)^n]dx

    I have to find a reduction formula for this question, but I've tried integrating by parts and trig substitution to no avail. When I integrate by parts, I just get an x with an increasing power for every repetition in the integral, and trig substitution has just been a disaster so far.

    If anybody knows how I could do these without killing another forest for paper that would be great.

    Thank you
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  2. #2
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     \text{Bear in mind that} \frac{d}{dx} \arctan(x) = \frac{1}{x^2+1}

    =  \displaystyle \int {x \arctan^2(x)\, dx}

     = \displaystyle \bigg[\frac{x^2}{2} \arctan^2(x)\bigg] - \int {\frac{x^2}{2} \times 2 \arctan(x) \times \frac{1}{x^2+1}\, dx}

    =  \displaystyle \bigg[\frac{x^2}{2} \arctan^2(x)\bigg] - \int {\frac{x^2}{x^2+1} \arctan(x) \, dx}

    =  \displaystyle \bigg[\frac{x^2}{2} \arctan^2(x)\bigg] - \int {\bigg(\frac{x^2 + 1 }{x^2+1}-\frac{1}{x^2+1}\bigg) \arctan(x) \, dx}

    =  \displaystyle \bigg[\frac{x^2}{2} \arctan^2(x)\bigg] - \int {\bigg(1-\frac{1}{x^2+1}\bigg) \arctan(x) \, dx}

    =  \displaystyle \bigg[\frac{x^2}{2} \arctan^2(x)\bigg] - \bigg[\bigg(x-\arctan(x)\bigg)\bigg(\arctan(x)\bigg)\bigg] - \int {\bigg(x-\arctan(x)\bigg)\frac{1}{x^2+1} \, dx}

    =  \displaystyle \bigg[\frac{x^2}{2} \arctan^2(x)\bigg] - \bigg[\bigg(x-\arctan(x)\bigg)\bigg(\arctan(x)\bigg)\bigg] - \int {\bigg(\frac{x}{x^2+1}-\frac{\arctan(x)}{x^2+1}\bigg) \, dx}

    You now have one more application of integration to go in the last term. These two can be done by substitution by recognising that the denominator is the differential of the numerator in the first term, and in the second term you are multiply  \arctan(x) its derivative!

    Final answer should be, after simplification:

    =  \displaystyle \frac{x^2}{2} \arctan^2(x) + \arctan^2(x) - x\arctan(x) - \frac{1}{2} \ln|x^2+1| +\frac{1}{2}\arctan^2(x) +C

    =  \displaystyle  \arctan(x)\bigg(\frac{x^2}{2}+\arctan(x) - x  +\frac{1}{2}\arctan(x)\bigg)- \frac{1}{2} \ln|x^2+1| +C

    =  \displaystyle  \arctan(x)\bigg(\frac{x^2}{2} - x  +\frac{3}{2}\arctan(x)\bigg)- \frac{1}{2} \ln|x^2+1| +C
    Last edited by Mush; January 23rd 2009 at 02:45 PM.
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  3. #3
    MHF Contributor red_dog's Avatar
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    1)

    \int x\arctan^2xdx=\int\left(\frac{x^2}{2}\right)'\arct  an^2xdx=

    =\frac{x^2}{2}\arctan^2x-\int \arctan x\frac{x^2}{x^2+1}dx=\frac{x^2}{2}\arctan^2x-\int\arctan x\frac{x^2+1-1}{x^2+1}dx=

    =\frac{x^2}{2}\arctan^2x-\int\arctan xdx+\int\arctan x\frac{1}{x^2+1}dx=

    =\frac{x^2}{2}\arctan^2x-\int x'\arctan xdx+\frac{\arctan^2x}{2}=

    =\frac{x^2}{2}\arctan^2x+\frac{\arctan^2x}{2}-x\arctan x+\int\frac{x}{x^2+1}dx=

    =\frac{x^2}{2}\arctan^2x+\frac{\arctan^2x}{2}-x\arctan x+\frac{1}{2}\ln (x^2+1)+C

    EDIT: You're right, Mush. I've edited.
    Last edited by red_dog; January 23rd 2009 at 12:52 PM.
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  4. #4
    MHF Contributor red_dog's Avatar
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    2)

    I_n=\int(x^2+1)^ndx=\int(x^2+1)^{n-1}(x^2+1)dx=

    =\int(x^2+1)^{n-1}\cdot x^2dx+I_{n-1}=\frac{1}{2n}\int x((x^2+1)^n)'dx+I_{n-1}=

    =\frac{1}{2n}x(x^2+1)^n-\frac{1}{2n}I_n+I_{n-1}

    Then, I_n=\frac{x(x^2+1)^n}{2n+1}+\frac{2n}{2n+1}I_{n-1}
    Last edited by red_dog; January 23rd 2009 at 12:53 PM.
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  5. #5
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    Quote Originally Posted by red_dog View Post
    1)

    \int x\arctan^2xdx=\int\left(\frac{x^2}{2}\right)'\arct  an^2xdx=

    =\frac{x^2}{2}\arctan^2x-\int \arctan x\frac{x^2}{x^2+1}dx=\frac{x^2}{2}-\int\arctan x\frac{x^2+1-1}{x^2+1}dx=

    =\frac{x^2}{2}-\int\arctan xdx+\int\arctan x\frac{1}{x^2+1}dx=\frac{x^2}{2}-\int x'\arctan xdx+\frac{\arctan^2x}{2}=

    =\frac{x^2}{2}+\frac{\arctan^2x}{2}-x\arctan x+\int\frac{x}{x^2+1}dx=

    =\frac{x^2}{2}+\frac{\arctan^2x}{2}-x\arctan x+\frac{1}{2}\ln (x^2+1)+C
    I believe the first term of your result should be multiplied by  \arctan^2(x) .
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  6. #6
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    Hello, bnay!

    I think I've solved the first one . . .


    1)\;\;I \:=\:\int x\cdot(\arctan x)^2\cdot dx

    Let: \theta \:=\:\arctan x  \quad\Rightarrow\quad x \:=\:\tan\theta \quad\Rightarrow\quad dx \:=\:\sec^2\!\theta\,d\theta

    Substitute: . \int\tan\theta\cdot \theta^2\cdot(\sec^2\!\theta\,d\theta) \;=\;\int\theta^2\cdot(\tan\theta\sec^2\theta\,d\t  heta)

    . . By parts: . \begin{array}{ccccccc}u &=& \theta^2 & & dv &=&\tan\theta\sec^2\!\theta\,d\theta \\ du &=& 2\theta\,d\theta & & v &=&\frac{1}{2}\tan^2\!\theta \end{array}

    We have: . I \;=\;\tfrac{1}{2}\theta^2\tan^2\!\theta - \int\theta\tan^2\!\theta\,d\theta

    . . By parts: . \begin{array}{ccccccc}u &=& \theta & & dv &=&\tan^2\!\theta\,d\theta \\<br />
du &=& d\theta && v &=& \tan\theta - \theta\end{array} .*


    We have: . I \;=\;\tfrac{1}{2}\theta^2\tan^2\theta - \bigg[\theta(\tan\theta - \theta) - \int (\tan\theta - \theta)\,d\theta

    . . . . . . . . I \;=\;\tfrac{1}{2}\theta^2\tan^2\!\theta - \theta(\tan\theta-\theta) + \int(\tan\theta - \theta)\,d\theta

    . . . . . . . . I \;=\;\tfrac{1}{2}\theta^2\tan^2\!\theta - \theta\tan\theta + \theta^2 - \ln(\cos\theta) -\frac{1}{2}\theta^2 + C

    . . . . . . . . I \;=\;\tfrac{1}{2}\theta^2\tan^2\theta - \theta\tan\theta + \tfrac{1}{2}\theta^2 - \ln(\cos\theta) + C


    Back-substitote: . \theta \,=\,\arctan x

    I \;=\;\tfrac{1}{2}(\arctan x)^2\left[\tan(\arctan x)\right]^2 - (\arctan x)[\tan(\arctan x)] + \tfrac{1}{2}(\arctan x)^2 - \ln|\cos(\arctan x)| + C

    I \;=\;\tfrac{1}{2}(\arctan x)^2\cdot x^2 - (\arctan x)\cdot x + \tfrac{1}{2}(\arctan x)^2 + \tfrac{1}{2}\ln(1+x^2) + C .**

    \boxed{I \;=\;\tfrac{1}{2}x^2(\arctan x)^2 - x\arctan x + \tfrac{1}{2}(\arctan x)^2 + \tfrac{1}{2}\ln(1+ x^2) + C}


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


    * . \int\tan^2\!\theta\,d\theta \:=\:\int(\sec^2\!\theta-1)\,d\theta \:=\:\tan\theta - \theta


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


    ** . We have: . \ln|\cos(\arctan x)| .[1]

    Then: . \alpha\:=\:\arctan x \quad\Rightarrow\quad \tan\alpha \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}

    So \alpha is in a right triangle with: .  opp = x,\;adj = 1
    Using Pythagorus, we have: . hyp \:=\:\sqrt{1+x^2}
    Hence: . \cos\alpha \:=\:\frac{1}{\sqrt{1+x^2}}

    Then [1] becomes: . \ln\left(\frac{1}{\sqrt{1+x^2}}\right)\;=\;\ln\lef  t(1+x^2\right)^{-\frac{1}{2}} \;=\;-\tfrac{1}{2}\ln(1+x^2)


    I need a nap . . .
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  7. #7
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    Smile

    Quote Originally Posted by Mush View Post
    <br />
=  \displaystyle \frac{x^2}{2} \arctan^2(x) + \arctan^2(x) - x\arctan(x) - \frac{1}{2} \ln|x^2+1| +\frac{1}{2}\arctan(x) +C<br />
    The last term is \frac{1}{2}\arctan^2(x).
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  8. #8
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    Quote Originally Posted by courteous View Post
    The last term is \frac{1}{2}\arctan^2(x).
    Indeed! That's the kind of mistakes you make when you manipulate things mentally.
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  9. #9
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    Thank you all very much. This was all very helpful, and for some reason I never thought of adding 1 and subtracting it (but I think 5 hours of math will do that, lol0
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