whats the intgration of sin^3(4t)?

Question

thanks guys....your answers confirmed that what i was doing was correct...

Answer 1

I = ∫ sin ³ (4t) dt
I = ∫ sin(4t) (1 - cos ² (4t) ) dt
Let u = cos 4t
du = - 4 sin 4t dt
I = (-1/4) ∫ 1 - u ² du
I = (-1/4) u + (1/12) u ³ + C
I = (-1/4) cos 4t + (1/12) cos ³ 4t + C

Answer 2

∫ sin^3(4t) dt

let 4t = x

4dt = dx

dt = (1/4)dx

∫ sin^3(4t) dt = 1/4∫sin^3(x) dx

=> 1/4 ∫ sin x (sin^2(x) dx= 1/4∫ sin x(1- cos^2(x) dx

=>1/4∫[sinx - sin x cos^2(x)] dx

=>1/4∫sin x dx -1/4∫ cos^2(x)(-1)d(cosx) [since d(cosx) = -sinx]

=>1/4 ∫ sin x dx + 1/4∫ cos^2(x) d(cosx)

=>1/4(-cosx) + 1/4(cos^3(x)/3 + c

=>(1/12) cos^3(x) - (1/4) cos x + c

substituting back x = 4t

=>(1/12) cos^3(4t) - (1/4) cos (4t) + c

Answer 3

∫sin³(4t) dt

Rewrite as:
∫sin²(4t) sin(4t) dt

using sin² + cos² = 1
sin² = 1 - cos²
∫[1-cos²(4t)] sin(4t) dt

u = cos(4t)
du = - 4 sin(4t) dt
-1/[ 4 sin(4t) ] du = dt

making the substitution.
-(1/4)∫[1-u²] du

Integrating using power rule
-(1/4)[u-u³/3] + C
-1/4u + u³/12 + C
(-1/4)cos(4t)+cos³(4t)/12 + C
cos³(4t)/12 - cos(4t)/4+ C

Answer 4

integ sin^n(x) = -sin^(n-1)xcos(x)/n + (n-1)/n integ sin^(n-2)x
n=3
let 4t=k
4dt=dk
dt=dk/4
use the above formula with n=3 and divide the answer by 4.
=-sin^2(k)cos(k)/3 +(2/3)integ sin(k)dt
=-sin^2(k)cos(k) / 3 -(2/3)cos(k) +c
replace k by 4t and divide the answer by 4.

Answer 5

1/48*(cos(12t))-9cos(4t))