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# Urgent P3 Help!! watch

1. 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???!!
2. 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 = ..)
3. (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)
4. 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)
5. 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?
6. 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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Updated: June 16, 2004
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