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Integrating Using a Substituation

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Use the substitution ##u=\frac{\pi} {2}-x## evaluate the integral ##\int_0^\frac {\pi}{2} \frac {\sin x}{\cos x + \sin x}dx##.

    2. Relevant equations

    ##\cos (\frac {\pi}{2}-x)=\sin x##

    3. The attempt at a solution

    I start by plugging "u" into the equation making the function become

    ##\int_0^\frac {\pi}{2} \frac {\sin( \frac{\pi} {2}-x)}{\cos(\frac{\pi} {2}-x) + \sin (\frac{\pi} {2}-x)}##.

    I substitute the sin x for the ##\cos (\frac {\pi}{2}-x)## because I know they are equal.

    Then i have

    ##\int_0^\frac {\pi}{2} \frac {\sin (\frac{\pi} {2}-x)}{\sin (x) + \sin (\frac{\pi} {2}-x)}##

    What comes next is a mystery to me.
    I feel maybe I did not use the substitution in a correct way which is why am puzzled about were to go next.

    I know that the answer is pi/4, but I'm not sure how to actually get there with substitutions.
     

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    Last edited: Mar 1, 2015
  2. jcsd
  3. Mar 1, 2015 #2

    Mark44

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    I see two mistakes:
    1) You are missing dx in the integral above. With your substitution, what is du?
    2) It doesn't look like you have changed the limits of integration to account for the substtitution.
     
  4. Mar 1, 2015 #3

    ElijahRockers

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    I haven't worked out the problem, but you might try the Sum-to-Product identity for sin(a) + sin(b).
     
  5. Mar 1, 2015 #4

    ElijahRockers

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    Mark, the limits actually flip, but there's a negative from du. Flipping the limits back eliminates the negative.
     
  6. Mar 1, 2015 #5

    Mark44

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    I'm aware of that, but I'm not convinced that the OP is.
     
  7. Mar 1, 2015 #6
    I incidentally made that choice to leave those bounds. I now know that now though.
     
  8. Mar 2, 2015 #7
    so for
     
  9. Mar 2, 2015 #8
    What I have is
    ## \sin x+\sin (\frac{\pi}{2}-x)##

    using the Sum to Product Identity

    = ##\sqrt{ 2}\cos \frac {(\frac{\pi}{2}-x)}{2}##

    Putting that into our integral would be:

    ##\int \frac {\sin (\frac {\pi}{2}-x)}{\sqrt{ 2}\cos \frac {(\frac{\pi}{2}-x)}{2}}\ dx##

    =##\int \frac {\cos x}{\sqrt{ 2}\cos \frac {(\frac{\pi}{2}-x)}{2}}\ dx##

    but then im not sure about what my next substitution is....
     
  10. Mar 2, 2015 #9

    ehild

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    Try to use the identity ## \sin(x)+\cos(x)= \sqrt{2}\sin(x+\pi/4)## and substitute u=x+pi/4. You get an easy integral in terms of u.
     
  11. Mar 2, 2015 #10

    SammyS

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    Why stop there.

    It's also true that ##\displaystyle \cos(x) = \sin(\frac {\pi}{2}-x)\ ## .

    Make that substitution also.


    (It's integration by trickery.)
     
  12. Mar 5, 2015 #11

    SammyS

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    I was doubtful regarding how clear I was in my previous post, even though Elijah "Liked" the post.

    The above quoted posts both address evaluating the given definite integral in the usual way; that is, by using the Fund. Thm. of Calc. I , find an anti-derivative, then evaluate that at the upper lower limits of integration.

    The hint given, to use the substitution, ##\ u=\frac{\pi} {2}-x \ ##, doesn't particularly aid in that regard.

    What the instructor, or whomever posed the question, likely had in mind was the following.
    Using that substitution to gives the following result:$$ \int_0^\frac {\pi}{2} \frac {\sin x}{\cos x + \sin x}dx=\int_0^\frac {\pi}{2} \frac {\cos x}{\cos x + \sin x}dx \ $$​
    Adding the integral on the left with that on the right results in a very simple integrand as well as giving twice the desired result.
     
  13. Mar 5, 2015 #12

    ehild

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    Ingenious !!!
     
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