What an easy integration?

Question

intergral of (sinx)^4*(cosx)^4dx?

"*" means multiply...

plz show all work..

thx

Answer 1

your "easy" sounds like a joke....it's actually somewhat complicated....!

∫ sin^4x  cos^4x dx =

replace sin^4x with (1 - cos^2x)^2:

∫ (1 - cos^2x)^2  cos^4x dx =

expand it as:

∫ (1 - 2cos^2x + cos^4x) cos^4x dx =

∫ (cos^4x  - 2cos^6x + cos^8x) dx =

now, in order to reduce the cosine powers, use the half-angle identity:

cos^2(θ/2) = (1/2)(1 + cosθ) → cos^2x = (1/2)[1 + cos(2x)]

therefore:

cos^4x = (cos^2x)^2 = {(1/2)[1 + cos(2x)]}^2 = (1/4)[1 + cos(2x)]^2 =

(1/4)[1 + 2cos(2x) + cos^2(2x)] =

cos^6x = (cos^2x)^3 =  {(1/2)[1 + cos(2x)]}^3 = (1/8)[1 + cos(2x)]^3 =

(1/8)[1 + cos^3(2x) + 3cos(2x) + 3cos^2(2x)] =

cos^8x = (cos^4x)(cos^4x) = [(1/4)[1 + cos(2x)]^2][(1/4)[1 + cos(2x)]^2] =

(1/4)[1 + 2cos(2x) + cos^2(2x)](1/4)[1 + 2cos(2x) + cos^2(2x)] =

(1/16)[1 + 2cos(2x) + cos^2(2x)][1 + 2cos(2x) + cos^2(2x)] =

(1/16)[1 + 2cos(2x) + cos^2(2x) + 2cos(2x) + 4cos^2(2x) + 2cos^3(2x) +

cos^2(2x) + 2cos^3(2x) + cos^4(2x)] =

(1/16)[1 + 4cos(2x) + 6cos^2(2x) + 4cos^3(2x) + cos^4(2x)]

consequently the integrand becomes:

∫ (cos^4x  - 2cos^6x + cos^8x) dx =

∫ {(1/4)[1 + 2cos(2x) + cos^2(2x)]  - 2 (1/8)[1 + cos^3(2x) + 3cos(2x) + 3cos^2(2x)] + (1/16)[1 + 4cos(2x) + 6cos^2(2x) + 4cos^3(2x) + cos^4(2x)]} dx =

∫ {(1/4)[1 + 2cos(2x) + cos^2(2x)]  - (1/4)[1 + cos^3(2x) + 3cos(2x) + 3cos^2(2x)] +

(1/4)(1/4)[1 + 4cos(2x) + 6cos^2(2x) + 4cos^3(2x) + cos^4(2x)]} dx =

factoring out (1/4):

(1/4) ∫ {[1 + 2cos(2x) + cos^2(2x)]  - [1 + cos^3(2x) + 3cos(2x) + 3cos^2(2x)] +

(1/4)[1 + 4cos(2x) + 6cos^2(2x) + 4cos^3(2x) + cos^4(2x)]} dx =

(1/4) ∫ [1 + 2cos(2x) + cos^2(2x) - 1 - cos^3(2x) - 3cos(2x) - 3cos^2(2x) +

(1/4) + (4/4)cos(2x) + (6/4)cos^2(2x) + (4/4)cos^3(2x) + (1/4)cos^4(2x) dx =

simplifying and adding similar terms:

(1/4) ∫ [(1/4)cos^4(2x)  - (1/2)cos^2(2x) + (1/4)] dx =

in order to furtherly reduce cosine powers, use the mentioned half-angle identity again:

cos^2x = (1/2)[1 + cos(2x)] →

cos^2(2x) = (1/2)[1 + cos(4x)] →

cos^4(2x) = [cos^2(2x)]^2 = {(1/2)[1 + cos(4x)]}^2 = (1/4)[1 + 2cos(4x) + cos^2(4x)] =

so that the integrand becomes:

(1/4) ∫ [(1/4)cos^4(2x)  - (1/2) cos^2(2x) + (1/4)] dx =

(1/4) ∫ { (1/4){(1/4)[1 + 2cos(4x) + cos^2(4x)]}  - (1/2){(1/2)[1 + cos(4x)]}  + (1/4) } dx =

(1/4) ∫ { (1/4)(1/4)[1 + 2cos(4x) + cos^2(4x)]  - (1/4)[1 + cos(4x)] + (1/4) } dx =

factoring out (1/4) again,

(1/4)(1/4) ∫ {(1/4)[1 + 2cos(4x) + cos^2(4x)]  - [1 + cos(4x)] + 1 } dx =

(1/16) ∫ [(1/4) + (2/4)cos(4x) + (1/4)cos^2(4x)  - 1 - cos(4x) + 1] dx =

(1/16) ∫ [(1/4) + (1/2)cos(4x) + (1/4)cos^2(4x) - cos(4x)] dx =

(1/16) ∫ [(1/4) - (1/2)cos(4x) + (1/4)cos^2(4x)] dx =

similarly, replace cos^2(4x) with (1/2)[1 + cos(8x)], yielding:

(1/16) ∫ {(1/4) - (1/2)cos(4x) + (1/4)(1/2)[1 + cos(8x)]} dx =

(1/16) ∫ {(1/4) - (1/2)cos(4x) + (1/8)[1 + cos(8x)]} dx =

(1/16) ∫ [(1/4) - (1/2)cos(4x) + (1/8) + (1/8)cos(8x)] dx =

(1/16) ∫ [(3/8) - (1/2)cos(4x) + (1/8)cos(8x)] dx =

factoring out (1/2),

(1/16)(1/2) ∫ [(3/4) - cos(4x) + (1/4)cos(8x)] dx =

(1/32) ∫ [(3/4) - cos(4x) + (1/4)cos(8x)] dx =

break it up into:

(1/32) ∫ (3/4) dx - (1/32) ∫ cos(4x) dx + (1/32) ∫ (1/4)cos(8x) dx =

taking the constants out,

(1/32)(3/4) ∫ dx - (1/32) ∫ cos(4x) dx + (1/32)(1/4) ∫ cos(8x) dx =

(3/128) ∫ dx - (1/32) ∫ cos(4x) dx + (1/128) ∫ cos(8x) dx =

finally integrate it into:

(3/128) x - (1/32) [(1/4)sin(4x)] + (1/128) [(1/8)sin(8x)] + C

thus, in conclusion:

∫ sin^4x  cos^4x dx = (3/128) x - (1/128) sin(4x) + (1/1024) sin(8x) + C

I hope it helps...

Bye!

Answer 2

∫(sinx)^4*(cosx)^4dx=

(1/32)∫(2sin x cos x)^4 d(2x)=

(1/32)∫sin^4 (2x) d(2x)

substitute 2x =t and obtain( without (1/32) for simplity

∫sin^4 t dt=

∫sin^2 t (1-cos^2 t) dt=

∫sin^2 t dt -1/16∫sin^2 4t d(4t)

from here use sin^2 t = (1-cos 2t)/2

idem for sin^2 4t = (1-cos 8t)/2

Now you recalc and substitute back