Indicate one method to solve the integral of y=sin4x*cos6x?

The first step in evaluating the indefinite integral is to
transform the given product of trigonometric functions into a
sum.

We'll apply the
identity:

sin a * cos b =
[sin(a+b)+sin(a-b)]/2

We'll substitute a by 4x and b by
6x.

sin4x*cos6x =
[sin(4x+6x)+sin(4x-6x)]/2

sin4x*cos6x = (sin 10x)/2 - (sin
2x)/2

Now, we'll evaluate the
integral:

Int sin4x*cos6x dx = Int (sin 10x)dx/2 - Int (sin
2x)dx/2

Int (sin 10x)dx = -(cos 10x)/10 +
C

Int (sin2x)dx = -(cos 2x)/2 +
C

The indefinite integral of the given
trigonometric product is: Int sin4x*cos6x dx = -(cos 10x)/20 + (cos 2x)/4 +
C