Question

integrate (sin(3x))^4

Answer 1

int ((1-cos(6x))/2)^2*dx
int ((1 - 2*cos(6x) + cos(6x)^2)/4) dx
int 1/4 dx -int cos(6x)/2 dx + int ((1+cos(12x))/8)dx
x/4 - sin(6x)/12+x/8+sin(12x)/96+k ....

Answer 2

so what is the process to the answer...

Answer 3

@experimentX
(sin(3x))^4 = [(sin3x)^2]^2 = (2cos6xsin6x)^2 ...
how come it becomes ((1-cos(6x))/2)^2 ?

Answer 4

can you somehow draw it because this is very confusing to read

Answer 5

cos 2x = cos(x)^2 - sin(x)^2
or, cos 2x = 1- 2sin(x)^2
so sin(x)^2 = (1-cos2x)/2

Answer 6

how did you get the denominators to be 12 and 96 for the fractions

Answer 7

similar for 96