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

Answer 1

Evelyn Agard

#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#

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Answer 2

Evelyn Agard

To find the integral of ( x \sin(6x) ), you can use integration by parts method. Let ( u = x ) and ( dv = \sin(6x) , dx ). Then, differentiate ( u ) to get ( du = dx ) and integrate ( dv ) to get ( v = -\frac{1}{6} \cos(6x) ). Apply the integration by parts formula: (\int u , dv = uv - \int v , du ). Substituting the values, you get: [ \int x \sin(6x) , dx = -\frac{1}{6}x\cos(6x) - \int (-\frac{1}{6}\cos(6x)) , dx ] Simplify the integral on the right side and integrate to get: [ \int x \sin(6x) , dx = -\frac{1}{6}x\cos(6x) + \frac{1}{36}\sin(6x) + C ] where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.