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How do you integrate #int xcos(2x)# by integration by parts method?

Answer 1

Ezra Adams

#1/4(2x*sin2x+cos2x)+C#.

Let #I=intxcos2xdx#

We use the Rule of Integration by Parts, which is,

#intuvdx=uintvdx-int[(du)/dxintvdx]dx#

We take,

#u=x, so (du)/dx=1; & v=cos2x, so, intvdx=(sin2x)/2#.

Hence, #I=x/2*sin2x-1/2intsin2xdx#

#=x/2*sin2x-1/2((-cos2x)/2)#

#=1/4(2x*sin2x+cos2x)+C#.

Enjoy Maths.!

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Answer 2

Emma Aldridge

To integrate ( \int x \cos(2x) ) using integration by parts method, you would use the formula:

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

where ( u ) and ( dv ) are chosen such that it simplifies the integral on the right side. Here, we can choose ( u = x ) and ( dv = \cos(2x) , dx ).

Then, we find ( du ) and ( v ) by differentiating ( u ) and integrating ( dv ):

[ du = dx ] [ v = \frac{1}{2} \sin(2x) ]

Now, we plug these into the integration by parts formula:

[ \int x \cos(2x) , dx = x \left( \frac{1}{2} \sin(2x) \right) - \int \left( \frac{1}{2} \sin(2x) \right) , dx ]

[ = \frac{1}{2}x \sin(2x) + \frac{1}{4} \cos(2x) + 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.