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# integration watch

1. 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
2. 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
3. note that sinx=cos((pi/2)-x)=cos(2((pi/4)-x/2))
then use the double angle formula for cos (2x).

tatum
4. (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
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)
5. (Original post by IntegralAnomaly)
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
6. (Original post by IntegralAnomaly)
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?
7. (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)
8. (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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