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Find the integral of the function y=sinx*cos(3x)*cos^2 x-cos(3x)*sin^3 x?
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First,we'll factorize by sin x*cos 3x and we'll get:
y = sin x*cos 3x*[(cos x)^2 - (sin x)^2]
We recognize the double angle formula:
[(cos x)^2 - (sin x)^2] = cos 2x
The function will become: y = sin x*cos 2x*cos 3x
Now, 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
Based on fact that cosine function is even, we'll replace 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 transform 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 the 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
The integral of the given function y is: Int ydx = (-cos 6x)/24 + (cos 4x)/16 - (cos 2x)/8 + C
Posted by giorgiana1976 on April 15, 2011 at 3:00 AM (Answer #1)
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