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    Iv got my exam in 2 hours and im majorly panicing... iv just gone blank...

    Iv forgotten how to integrate... how do you integrate sin^3x and sin^4x???!!
    My eternal gratitude to anyone who can help!! please please please!!
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    sin^3 x = sinx(1 - cos^2 x) you can use a substitution of u = cosx for the second bit or just notice that it's the derivative of 1/3*cos^3 x

    sin^4 x = sin^2 x(1 - cos^2 x) = sin^2 x - 1/2*sin^2 (2x)
    Then use cos(2x) = 1 - 2sin^2 x for both bits (second part you use cos4x = ..)
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    (Original post by Bezza)
    sin^3 x = sinx(1 - cos^2 x) you can use a substitution of u = cosx for the second bit or just notice that it's the derivative of 1/3*cos^3 x

    sin^4 x = sin^2 x(1 - cos^2 x) = sin^2 x - 1/2*sin^2 (2x)
    Then use cos(2x) = 1 - 2sin^2 x for both bits (second part you use cos4x = ..)
    right thanks!! im still a bit confused though... i thought you couldn't integrate powers of sin and cos straight out...
    (Original post by Bezza)
    sin^4 x = sin^2 x(1 - cos^2 x) = sin^2 x - 1/2*sin^2 (2x)
    :confused:
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    cos(2x) = 1 - 2sin^2 x
    sin^2 x = (1 - cos(2x))/2
    so integral of sin^2 x = x/2 - 1/4*sin(2x)
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    Im so confused... this is how i thought id work out $(sin^3x) ($ = integral)

    sin^3x = sin^2x*sinx (sin^2x = 0.5(1-cos2x))
    0.5$(sinx)(1-cos2x)
    = 0.5$sinx - cos2xsinx

    u=cos2x du/dx=-2sin2x
    dv/dx=sinx v=-cosx

    1/2[-cosx] - cos^2(2x) - $(2sin2xcosx)

    i can't see what im doing wrong... can you stop it? :confused:
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    You only want to use the double angle cosine formula when it actually comes to integrating sin^2 x or cos^2 x. For simplify sin^3 x it's easiest to say sin^2 x = 1 - cos^2 x

    $ sin^3 x dx = $ sinx(1 - cos^2 x) dx = $ (sinx - sinx*cos^2 x) dx

    Then use substitution instead of parts, u = cosx, du/dx = -sinx
    so $ sin^3 x dx = $ sinx dx + $ u^2 du = -cosx + 1/3*u^3 + c = -cosx + 1/3*cos^3 x + c
 
 
 

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