What is the antiderivative of y=sinx*cos2x*cos3x?

giorgiana1976 | Student

First, we'll transform the product cos2x*cos3x into a sum:

cos2x*cos3x = [cos(2x + 3x) + cos (2x - 3x)]/2

cos2x*cos3x = [cos(5x) + cos (-x)]/2

Since cosine function is even, we'll put cos(-x) = cos x.

cos2x*cos3x = (cos 5x+ cos x)/2

The function will become:

y = sin x*cos 5x/2 + sin x*cos x/2

We'll transfomr the product sin x*cos 5x into a difference:

sin x*cos 5x = [sin(x+5x) + sin(x-5x)]/2

sin x*cos 5x = [sin(6x) + sin(-4x)]/2

Since sine function is odd, we'll put sin(-x) = -sin x.

sin x*cos 5x = (sin 6x - sin 4x)/2

The 2nd term of y is sin x*cos x/2 = 2*sin x*cos x/2*2 = sin 2x/4

We'll evaluate the integral of te function:

Int ydx = Int (sin 6x - sin 4x)dx/4 + Int sin 2xdx/4

Int ydx = Int (sin 6x)dx/4 - Int(sin 4x)dx/4 + Int sin 2xdx/4

Int ydx = -cos 6x/24 + cos 4x/16 - cos 2x/8 + C

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