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    i 4got how d hel 2 do dis!!!

    integrate sqrt(1+sin x) its got sutin 2 do wit half angle formula but i cant get it
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    do you mean the t = tan (x/2) substitution?

    if so, sin(x) = 2t/(1+t²) and dx = 2/(1+t²) dt

    => Int sqrt(1+sin(x)) dx = Int sqrt(1+2t/(1+t²)) 2/(1+t²) dt
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    note that sinx=cos((pi/2)-x)=cos(2((pi/4)-x/2))
    then use the double angle formula for cos (2x).

    tatum
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    (Original post by chetan)
    i 4got how d hel 2 do dis!!!

    integrate sqrt(1+sin x) its got sutin 2 do wit half angle formula but i cant get it
    how about this 1:
    1+sinx=cos² (x/2)+sin² (x/2)+2sin(x/2)cos(x/2)
    (1=cos² (x/2)+sin² (x/2) and sin(x) in terms of half angles is 2sin(x/2)cos(x/2))
    so 1+sinx=[cos(x/2)+sin(x/2)]²
    so √(1+sinx)=cos(x/2)+sin(x/2)
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    (Original post by IntegralAnomaly)
    how about this 1:
    1+sinx=cos² (x/2)+sin² (x/2)+2sin(x/2)cos(x/2)
    (1=cos² (x/2)+sin² (x/2) and sin(x) in terms of half angles is 2sin(x/2)cos(x/2))
    so 1+sinx=[cos(x/2)+sin(x/2)]²
    so √(1+sinx)=cos(x/2)+sin(x/2)
    Neat
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    (Original post by IntegralAnomaly)
    how about this 1:
    1+sinx=cos² (x/2)+sin² (x/2)+2sin(x/2)cos(x/2)
    (1=cos² (x/2)+sin² (x/2) and sin(x) in terms of half angles is 2sin(x/2)cos(x/2))
    so 1+sinx=[cos(x/2)+sin(x/2)]²
    so √(1+sinx)=cos(x/2)+sin(x/2)
    Someone talk me through this?
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    (Original post by ZJuwelH)
    Someone talk me through this?
    ok u know that sin(2x)=2sinxcosx right?
    so sinx=2sin(x/2)cos(x/2).All i've done is swapped x by x/2.
    and u know that sin²x+cos²x≡1 yes?
    so again swappin x with x/2 sin²(x/2)+cos²(x/2)≡1.
    so 1+sinx=cos²(x/2)+2sin(x/2)cos(x/2)+sin²(x/2)=[cos(x/2)+sin(x/2)]²
    (just factorised the expression)
    so by takin the square root u get:
    √(1+sinx)=cos(x/2)+sin(x/2)
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    (Original post by IntegralAnomaly)
    ok u know that sin(2x)=2sinxcosx right?
    so sinx=2sin(x/2)cos(x/2).All i've done is swapped x by x/2.
    and u know that sin²x+cos²x≡1 yes?
    so again swappin x with x/2 sin²(x/2)+cos²(x/2)≡1.
    so 1+sinx=cos²(x/2)+2sin(x/2)cos(x/2)+sin²(x/2)=[cos(x/2)+sin(x/2)]²
    (just factorised the expression)
    so by takin the square root u get:
    √(1+sinx)=cos(x/2)+sin(x/2)
    I like that a lot, nice one, thanks.
 
 
 

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