Question

find integral of -6
sin(3x) dx​

Answer 1

Answer:

-6 of (-)cos3x/3+constant

Step-by-step explanation:

sin 3x=sin(2x+x)=sin 2x cos x+cos 2x sin x. So, Intsin 3x dx=I (say)

I= Int sin 2x cos x dx+Int cos 2x sin x dx

or, I = sin2x Intcos x dx-Int[d/dx sin 2x(intcos xdx) dx]+cos 2x int sinxdx-Int[d/dxcos 2x(intsin xdx)dx]

or,I=sin 2x*sin x-2*Intcos 2x*sinx dx+cos 2x(- cos x)+2*Int sin 2x(-cos x)dx

or, I=-(cos 2x*cos x-sin 2x*sinx)-2Int(cos 2x sinx+sin2x cosx)dx

or, I=-cos(2x+x)-2*I, or, 3I=-cos 3x, or, I= (-)cos3x/3+constant

Here Int means integer