Question

How do I evaluate the indefinite integral `intsin(3x)*sin(6x)dx` ?

Answer 1

It can have two solutions

`=(sin3x)/6-(sin9x)/18+c`, where `c` is a constant OR

`=2/9sin^3(3x)+c`, where `c` is a constant

Explanation

`=intsin(3x)*sin(6x)dx`

From trigonometric identities,

`sinAsinB=1/2(cos(A-B)-cos(A+B))`

Similarly, for the given problem,

`sin(3x)*sin(6x)=1/2(cos(-3x)-cos9x)`

As, `cos(-A)=cos(A)`

`sin(3x)*sin(6x)=1/2(cos3x-cos9x)`

integrating both sides,

`intsin(3x)*sin(6x)dx=int1/2(cos3x-cos9x)dx`

`=1/2int(cos3x)dx-1/2int(cos9x)dx`

`=1/2(sin3x)/3-1/2(sin9x)/9+c`, where `c` is a constant

`=(sin3x)/6-(sin9x)/18+c`, where `c` is a constant

Another Method :

`=intsin(3x)*sin(6x)dx`

`=intsin(3x)*2sin(3x)cos(3x)dx`

`=int2sin^2(3x)cos(3x)dx`

let's assume `sin(3x)=t`, then `3cos(3x)dx=dt`

therefore,

`=2intt^2/3*dt`

`=2/9*t^3+c`, where `c` is a constant

`=2/9sin^3(3x)+c`, where `c` is a constant