Homework Help

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

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colorcolour | Student, College Freshman | eNoter

Posted December 28, 2010 at 5:39 PM via web

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

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 28, 2010 at 6:05 PM (Answer #2)

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To find the area under the curve y = cos x*sin 5x and between the lines x = 0 and x= pi, is the definite integral of y between x = 0 and x = pi.

Now Int [y] = Int [ cos x*sin 5x dx]

=> - (cos 6x)/12 - (cos 4x)/8 + C

At x = 0:

- (cos 6x)/12 - (cos 4x)/8 + C

= - 1/12 - 1/8 + C

At x = pi

= +1/12 + 1/8 + C

Subtracting the two we get 1/12 + 1/8 + C + 1/12 + 1/8 - C

=> 2/12 + 2/8

=> 5/12

Therefore the required area is 5/12

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neela | High School Teacher | Valedictorian

Posted December 28, 2010 at 6:05 PM (Answer #3)

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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.

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