How do you find #int xsin(6x) #?

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Answer 1 · 2015-11-21T12:16:02

#1 / 36 sin 6x-1/6 x cos 6x#

we want to disappear with "x" factor.

#u = x, du = dx#

#dv = sin 6x Rightarrow v = int sin 6x text{ d}x#

#A = int u text{ d}v = xv - int v text{ d} x#

#w = 6x Rightarrow frac{dw}{6} = dx#

#v = int sin w (dw)/6 = - 1/6 cos w#

#A = -x/6 cos 6x + int 1/6 (cos w) (dw)/6#

#A = -x/6 cos 6x + 1 / 36 sin w#