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  • integrate
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  • I have a solution ,

    I = ∫ 2 sin(x)/3 + sin(2x)

    Proceed by Integration by parts

    f1 = 2sinx
    f2 = 1/3 + sin(2x)

    I = f2*∫f1 - ∫(f2'*∫f1)
    = -2cosx/3 + sin(2x) + ∫4cos2x*cosx/(3 + sin2x)^2

    We Integrate again by parts

    I1 = ∫4cos2x*cosx/(3 + sin2x)^2
    This time ,
    f1 = 4cos2x/(3+sin2x)^2
    f2 = cosx

    I1 = -2cosx/ (3 + sin2x) - ∫2sinx/(3+sin2x)

    I1 = -2cosx/ (3 + sin2x) -I
    Also ,

    I1 =I +2cosx/3 + sin(2x)

    So Find I......

    I hope the solution works for you !! 8)
  • why is f1 taken as 2sin(x) and f2 as 1/3+sin(2x)
  • Gud question .... But always keep in mind that when you are using integration by parts,
    you will have to break the original function in such a way that
    one of them is conveniently integrable and the other differentiable.

    In this case : 2sin(x) is integrable as well as differentiable
    but , 1/3+sin(2x) cannot be integrated ....

    So , we break the function and integrate 2sin(x) and differentiate 1/3 + sin(2x) and proceed...

    Does this answer your question ???

    Always remember ... when a question of integration is asked in IIT-JEE / AIEEE or any other engineering examination ... There is a solution ...
    It is just about thinking in such a way that you get there easily and quickly !!!!
  • Neways ... We have been corresponding for a while now ...

    Which class are you in ??? I am a final year student at IIT kharagpur ..

    How are your prepations going on ???
    Always feel free to ask any question or send me a private message ....

    Best of Luck !!
  • wow...what a great TIP Friction!!! Gonna keep that in mind when I apper for the JEE...thx a lot!! :-)

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