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    integrate 2*pi*sinx*(1+(cosx)^2)^(1/2), between the interval 0<x<pi
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    (Original post by Simbo)
    integrate 2*pi*sinx*(1+(cosx)^2)^(1/2), between the interval 0<x<pi
    use the substitution cosx=u then ur integrand should become (1+u^2)^(1/2),then use the substitution u = sinht
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    Crikey.

    2π sinx √(1+cos²x)

    That's a tough one
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    thats in edexcel p5
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    yeah, ive never done edexcel p5...i dont get the second bit, where does sinht come from?
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    (Original post by Simbo)
    yeah, ive never done edexcel p5...i dont get the second bit, where does sinht come from?
    u've never done p5? so u've never herd of the hyperbolic function,which sinht is ofcourse
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    obviously not...i have now tho
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    (Original post by Simbo)
    obviously not...i have now tho
    thats strange,what syllabus are u doin? u should'nt be asked to integrate that without first coming across the hyperbolic functions
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    here hope dis helps!
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    thanx for that
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    (Original post by Simbo)
    integrate 2*pi*sinx*(1+(cosx)^2)^(1/2), between the interval 0<x<pi
    Its just the integral of 2 x pi x sinx^2. (use identity 1 + cos^2 = sin^2, root that to get sin)
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    yep it does, thanx alot
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    (Original post by mast3486)
    Its just the integral of 2 x pi x sinx^2. (use identity 1 + cos^2 = sin^2, root that to get sin) To get this integral you do it by parts or by an appropriate reduction formula (id recommend reduction formula for sin^n and then you can do it for any power of sin)
    Hope this helps
    woops! no you don't! That identity is wrong! should be sin^2 = 1 - cos^2, sorry i was thinkin of hyperbolic: cosh^2 - sinh^2 = 1) Doh!
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    dont worry, thanx anyway
 
 
 

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