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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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denyss | Student, College Freshman | eNoter

Posted April 15, 2011 at 2:52 AM via web

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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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giorgiana1976 | College Teacher | Valedictorian

Posted April 15, 2011 at 3:00 AM (Answer #1)

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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

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