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Basic integration

  1. Jun 16, 2017 #1
    1. The problem statement, all variables and given/known data

    Integrate: (sinx + cosx)/sqrt(1+sin2x)
    2. Relevant equations
    Simple trigo formulae:
    cos2x=cos^2(x)--sin^2(x)
    sin^2(x)+cos^2(x)=1
    sin2x=2.sinx.cosx
    3. The attempt at a solution
    I tried to rationalize the given term,multiplying both numerator and denominator with:
    1st time:cosx-sinx
    2nd: sqrt(1-sin2x)
    3rd: Both the above terms
    Everytime,I only got slightly different terms.

    So,went to substitution.
    But,I'm not finding any suitable term which I can substitute.
    If after rationalizing,I substitute sin2x as a variable,say,T,things might have worked out,
    EXCEPT that since I multiplied the term both to numerator and denominator,the other term just makes it impossible to express the whole term in terms of a simple integral.
    Any hints? (With or without substitution)
     
  2. jcsd
  3. Jun 16, 2017 #2

    Nidum

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    Do you know about integration by parts and quotient rule integration by parts ?
     
    Last edited: Jun 16, 2017
  4. Jun 16, 2017 #3
    ##(\cos x + \sin x)^2 = ?##
     
  5. Jun 16, 2017 #4
    By parts?Yeah,Quotient rule FOR INTEGRATION?I don't think so
     
  6. Jun 16, 2017 #5
    1+sin2x,I know,but I already told ya
    Then,I have a (sinx + cosx) in the denominator as well,right?That's my problem.
    I tried that way as well.
     
  7. Jun 16, 2017 #6

    Ray Vickson

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    If you write 1 as ##\sin^2 x + \cos^2 x##, the denominator becomes
    $$\sqrt{ \sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{(\sin x + \cos x)^2}.$$
    Can you see how to simplify ##\sqrt{(\sin x + \cos x)^2}\:?##
     
  8. Jun 16, 2017 #7
    Ooo...should have thought of that..
    K,got it.Thanks!
     
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