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Reduction formula problem

  1. May 12, 2014 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    Here is my attempt :

    19axlf.jpg


    I just can't get the answer the writing near the bottom is me trying to get the left hand side , so you can basically ignore that. But I don't think I made the correct first step... any advice?

    Didn't notice the picture was so blurry, the first thing I did was just proceed to integrate In by parts.
     

    Attached Files:

  2. jcsd
  3. May 13, 2014 #2

    BiGyElLoWhAt

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    Gold Member

    Yea your parts wasn't done properly, check it:

    ##\int_{0}^{\frac{\pi}{2}} e^{2x}sin^nx ##
    gotta do a parts:
    ##u = e^{2x}##
    ##dv = sin^nxdx##
    ##du=2e^{2x}##
    ##v = ...##
    gotta do another parts and use ##sin^2 + cos^2 =1##

    ##\int_{a}^bsin^nxdx = \int_{a}^b[sin^{n-2}(x)][1-cos^2(x)]dx =
    \int_{a}^b sin^{n-2}(x)-\int_{a}^{b} sin^{n-2}(x)cos(x)cos(x)## ...
    this just ends up at the reduction formula, and I don't feel like going all the way there, 2 more parts then nest them back together, but the point is: once you get your v
    plug it back into the first set of parts and you'll get something really messy that you need to tidy up. I bet that'll simplify to what you have going on in your homework problem.

    PS I can't really read the picture with your work
     
    Last edited: May 13, 2014
  4. May 13, 2014 #3
    I did not think of separating it then making subbing sin^2 x . Thanks.. I'm such an idiot. I just proceeded to integrate by parts without calling sin^n x , sin^(n-2) x* sin^2x
     
  5. May 13, 2014 #4
    I still don't get you fully, here is my working : yyli8.jpg

    Ignore the first line. where I said In = e^2x sinx^(n-1) * sin^x

    I also can't really do parts by the u, dv way or whatever. My teacher said he didn't want me to learn it like that. I just know I should always try to integrate the easier one first and differentiate the other one..
     
    Last edited: May 13, 2014
  6. May 13, 2014 #5

    SammyS

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    What do you mean by
    "I also can't really do parts by the u, dv way or whatever." ?


    For the second integration it looks like you have:
    ##u=\sin^{(n-1)}(x)\,\cos(x)\ ## and ##\ dv=e^{2x}dx \ .##​
    Those are good choices.

    What is the derivative of
    ##u=\sin^{(n-1)}(x)\,\cos(x)\ \ ?##​

    It's not what you have.
     
  7. May 14, 2014 #6
    I mean when I carry out integration by parts, I don't call things u or dv or v. I just proceed to integrate. That is how I was taught. The derivative of that would be (n-1)sin^(n-2)xcos^2x -sin^n
     
  8. May 15, 2014 #7

    SammyS

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    A few parentheses would help here making that derivative:
    (n-1)sin(n-2)(x)cos2(x) -sinn(x) .

    Notice that cos2(x) = 1 - sin2(x) .

    And as you say "integrate the easy part": [itex]\displaystyle \int e^{2x}dx[/itex]

    Integration by parts for the integral on your second line:
    [itex]\displaystyle \int e^{2x}\sin^{n-1}x \cdot \cos x\,dx[/itex]​

    will give you an integrand with two terms after you simplify.

    One of the integrals can be written as In , the other as In-2 .
     
  9. May 15, 2014 #8
    I got it out finally, I made a mistake with brackets at the 2nd parts... -_-
     
  10. May 15, 2014 #9
    Thank you guys.
     
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