intergration

Question

intergrate the following question:  sin(3x+a).sin(x-a)

Answer 1

Question: intergrate the following question:  sin(3x+a).sin(x-a).

There is a formula for the product of two sines.

sinA.sinB = (1/2)(cos(A-B) - cos(A+B))

Using this formula gives us,

I = int sin(3x+a).sin(x-a) dx = int (1/2)(cos((3x+a) - (x-a)) - cos((3x+a) + (x-a))) dx

I = int (1/2)(cos(2x+2a) - cos(4x)) dx = (1/2)((1/2)sin(2x+2a) - (1/4)sin(4x))

Answer: I = (1/4).sin(2x+2a) - (1/8).sin(4x)

Answer 2

∫sin(3x + a).sin(x -a) dx

let......... I =∫sin(3x + a).sin(x -a) dx

let...... u = sin(3x - a)............ du/dx = 3cos(3x + a)

dv =sin(x - a) dx................ v = -cos(x - a)

I = -cos(x -a).sin(3x - a) + 3∫cos(3x + a)cos(x  -a) dx

Again

let........ u = cos(3x + a)................ du/dx = -3sin(3x + a)

dv = cos(x - a) dx................. v = sin(x - a)

∫cos(3x + a)cos(x  -a) dx = cos(3x + a)sin(x - a) +3∫sin(3x + a)sin(x - a)

                                       =cos(3x + a)sin(x - a) + 3I

I = -cos(x - a)sin(3x + a) + 3{cos(3x + a)sin(x - a)} + 9I

I = 1/8{cos(x - a)sin(3x + a) -3cos(3x + a)sin(x - a)}

=by expressing in the form of sinP + sinQ = 2sin1/2(P + Q)cos1/2(P - Q)  and 

sinP - sinQ =2cos1/2(P + Q)sin(P -Q)

∫sin(3x + a).sin(x -a) dx=1/4sin(2x + 2a) - 1/8sin4x + c