Find the area under the curve y=cosx*sin5x and between the lines x=0 and x=pi.

To find the area under y=cosx*sin5x and between the lines
x=0 and x=pi.

We know that 2sinA*cosB = {sin(A+B) -+sin
(A-B)}

Therefore cosx*sin5x = sin5x*cosx =  (1/2)
{sin(5x+x)/2-sin(5x-x)/2} = (1/2) {sin3x- sin2x).

Therefore
Area under the curve  y = cosx*sin5x is Integral y dx Integral (sin5x*cosx) dx = Int
(1/2){sin3x-sin2x} dx, from x= 0 to x= pi.

 = (1/2) Int
{--(sin3x)+sin2x} dx from x= 0 to x = pi.

Let F(x) = -(1/2)
{ (cos3x)/3 + cos2x)/2}.

F(pi) = -{(1/2) {(cos3pi)/3 +
(cos2pi)/2} = - (1/2){-1/3+1/2} = -1/3.

F(0) =
(-1/2){1/3+1/2} = --5/12.

Therefore Area = F(pi)-F(0) =
-1/3-(-5/12) = (5-4)/12 = 1.

Therefore the area under y =
cox*sin5x from x= 0 to pi is 1/12 sq units.