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How do you evaluate the integral #int xsin(3x^2+1)#?

Answer 1

Isabella Ackerman

#intxsin(3x^2+1)dx=-1/6cos(3x^2+1)+C#

#I=intxsin(3x^2+1)dx#

Let #u=3x^2+1#, so #du=6xcolor(white).dx#.

#I=1/6intsin(3x^2+1)(6xcolor(white).dx)#

#I=1/6intsin(u)du#

#I=-1/6cos(u)+C#

#I=-1/6cos(3x^2+1)+C#

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Answer 2

Isaiah Allen

To evaluate the integral ( \int x \sin(3x^2 + 1) ), you can use the technique of integration by parts.

Integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Let ( u = x ) and ( dv = \sin(3x^2 + 1) , dx ). Then, differentiate ( u ) to find ( du ), and integrate ( dv ) to find ( v ).

[ du = dx ] [ v = -\frac{1}{3} \cos(3x^2 + 1) ]

Now, apply the integration by parts formula:

[ \int x \sin(3x^2 + 1) , dx = -\frac{x}{3} \cos(3x^2 + 1) - \int -\frac{1}{3} \cos(3x^2 + 1) , dx ]

[ = -\frac{x}{3} \cos(3x^2 + 1) + \frac{1}{9} \sin(3x^2 + 1) + 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.