Int(sin(2x)^3 cos(2x)^4) ?

Question

integral of sin^3 (2x) times cos^4 (2x)

Answer 1

∫ sin^3(2x) cos^4(2x) dx =

since sine is odd-powered, rearrange the integrand as:

∫ sin^2(2x) cos^4(2x) sin(2x) dx =

and rewrite the first factor in terms of cos(2x) as:

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

expand the part in terms of cos(2x) into:

∫ [cos^4(2x) - cos^6(2x)] sin(2x) dx =

let cos(2x) = u

differentiate both sides into:

d[cos(2x)] = du →

2 [-sin(2x)] dx = du →

sin(2x) dx (that is, what actually appears in your integral) = (-1/2) du

thus, substituting, you get:

∫ [cos^4(2x) - cos^6(2x)] sin(2x) dx = ∫ (u^4 - u^6) (-1/2) du =

(-1/2) ∫ (u^4 - u^6) du =

(-1/2) ∫ u^4 du + (1/2) ∫ u^6 du =

(-1/2) [u^(4+1)] /(4+1) + (1/2) [u^(6+1)] /(6+1) + C =

(-1/2)(1/5) u^5 + (1/2)(1/7) u^7 + C =

(-1/10) u^5 + (1/14) u^7 + C

then substitute back u = cos(2x), yielding:

∫ sin^3(2x) cos^4(2x) dx = (-1/10) cos^5(2x)+ (1/14) cos^7(2x) + C

I hope this helps..

Bye!

Answer 2

when ever there is cos*sin problem choose the even function to be t

let cos(2x)=t

-2sin(2x)dx=dt

now integral wud be

int((-1/2)(1-cos^2(2x)*cos^4(2x)*sin(2... )

= int ((-1/2)(1-t^2)(t^4))dt

= -1/2(t^5/5+ t^7/7)

now put t= cos(2x)

Answer 3

jhakaas dear, tumne last step me plus -minus ki mistake kar di.

∫(sin³2x)*(cos2x)^4 dx

∫sin²2x*sin2x*(cos2x)^4   dx

∫(1 - cos²2x)*(cos2x)^4*( sin2x) dx

Assume cos2x = u

-2sin2xdx  = du

sin2xdx  = -du/2

-1/2∫(1 - u²)*u^4 du

1/2∫u^6du -  1/2∫u^4du

(u^7)/14  - (u^5)/10  + c

[(cos2x)^7]/14  -  [(cos2x)^5]/10 + c