What is the integration of (sin (3x))^2 dx?

Question

The problem is written in the form sin(nt).sin(nt) dt

Answer 1

for the general case (where you have the n's) you need to integrate by parts twice then you reduce terms and you get your ansewr in terms of the n's. But for your case you just have to remember the identity:

(sin x)^2=1/2(1-cos(2x)).

Answer 2

Int (sin 3x)^2dx = x sin^2(3x) -Intx(2sin3xcos3x)*3 dx
The last

integral Is
Intx(sin6x)dx=(-cos6x)/6*x+1/6Intcos6xdx= 1/6(-x cos 6x+1/6sin6x)

Int = x*sin^2(3x)-1/2(-x*cos 6x+1/6 sin 6x)
All integrations performed by parts

Answer 3

Since cos2y = 2(siny)^2 - 1,
so cos2(3x) = 2(sin3x)^2 - 1
and (sin3x)^2 = (1/2)(cos6x + 1) = (1/2)cos6x + (1/2)

Integral of (sin3x)^2 dx
= Integral of (1/2)cos6x + (1/2) dx
= (1/2)sin6x.(1/6) + (1/2)x + C
= (1/12)sin6x + (1/2)x + C