How do I solve int(sin^5(2x)cos(2x)dx) ?

Answer 1

this is a u sub problem

let u = sin(2x) than du = 2 cos 2x dx and than du/2 = cos(2x) dx

now subbing it in we get

int(1/2 u^5 du)

which is 1/2 x 1/6 u^6 = 1/12 sin^6(2x) + c

Answer 2

the first guy is right but here is a little bit more detail if you want it.

int[ (sin^5 2x)(cos2x) dx]

let u=sin2x
then du/dx = d(sin2x)/dx
=> du/dx = (cos2x)(2) ... [d(sinx)/dx = cosx]
=>du/2 = (cos2x)(dx)

Put the value of u and du in the orig equation

int[ (sin^5 2x)(cos2x) dx]
=int [u^5 (1/2) (du)]
=1/2 int [u^5 du]
=(1/2)(1/6)u^6 + c

Substitute with the value of u

=1/12 (sin2x)^6 + c

That give you the answer

Answer 3

let 2x=y then dx=dy/2
1/2 (int) sin^5cosydy
let siny=t then cosydy=dt(by differentiating as above in step1
then you will be left with 1/2 int t^4dt
answer would be 1/10 sin^5
all the best

Answer 4

dhaval70 got it right.