If #f(x) = P sin(7x) + Qcos(7x)# and #f'(x) = xsin(7x)#, find #int_(9)^(pi//7) xsin(7x)dx#?

1 Answer
Truong-Son N.
Aug 13, 2017

I got

#int_(9)^(pi//7) xsin(7x) = 1/49[pi - sin(63) + 63cos(63)]#

Well, take the derivative so far...

#f'(x) = 7Pcos(7x) + Q(x cdot -7sin(7x) + cos(7x))#

#= 7Pcos(7x) - 7Qxsin(7x) + Qcos(7x)#

If we set #f'(x) = xsin(7x)#, then...

#(7P + Q)cos(7x) + (-7Q)xsin(7x) = 0cos(7x) + xsin(7x)#

In order for this equality to be satisfied, we have:

#7P + Q = 0#

#-7Q = 1#

Therefore, #ul(Q = -1/7)#, and...

#7P - 1/7 = 0#

#ul(P = 1/49)#

Knowing that, we have

#int_(9)^(pi//7) xsin(7x)dx = int_(9)^(pi//7) f'(x)dx = int_(9)^(pi//7) (df(x))/(dx)dx#

The integral over #x# of a #(df)/(dx)# derivative cancels it out.

#color(blue)(int_(9)^(pi//7) xsin(7x)dx) = |[stackrel(" ")stackrel(" ")(f(x))]|_(9)^(pi//7)#

#= |[1/49 sin(7x) - x/7 cos(7x)]|_(9)^(pi//7)#

#= [1/49 sin(7 cdot pi/7) - pi/49 cos(7cdot pi/7)] - [1/49 sin(7 cdot 9) - 9/7 cos(7 cdot 9)]#

#= [pi/49] - [1/49 sin(63) - 63/49 cos(63)]#

#= 1/49[pi - sin(63) + 63cos(63)]#

#~~ color(blue)(1.328)#

which is indeed what this integral is supposed to be.

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