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Urgent P3 Help!!

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!! :frown:
Reply 1
Avatar for Bezza
Bezza
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 = ..)
Reply 2
Avatar for chucky123
chucky123
OP
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!! :smile: im still a bit confused though... i thought you couldn't integrate powers of sin and cos straight out...
Bezza
sin^4 x = sin^2 x(1 - cos^2 x) = sin^2 x - 1/2*sin^2 (2x)
:confused:
Reply 3
Avatar for Bezza
Bezza
2
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)
Reply 4
Avatar for chucky123
chucky123
OP
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:
Reply 5
Avatar for Bezza
Bezza
2
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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