How do you integrate #int xsinxcosx# by integration by parts method?

Answer 1

The answer is #=(sin2x)/8-(xsin2x)/4+C#

We use #sin2x=2sinxcosx#
#intxsinxcosxdx=1/2intxsin2xdx#

The integration by parts is

#intuv'=uv-intu'v#
#u=x#, #=>#, #u'=1#
#v'=sin2x#, #=>#, #v=-(cos2x)/2#
so, #intxsin2xdx=-(xcos2x)/2+1/2intcos2xdx#
#=-(xcos2x)/2+1/2*(sin2x)/2#
#=(sin2x)/4-(xcos2x)/2#

And finally

#intxsinxcosxdx=1/2((sin2x)/4-(xcos2x)/2) +C#
#=(sin2x)/8-(xsin2x)/4+C#
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Answer 2

To integrate ( \int x \sin(x) \cos(x) ) using integration by parts:

Let ( u = x ) and ( dv = \sin(x) \cos(x) , dx ). Then ( du = dx ) and ( v = -\frac{1}{2} \sin^2(x) ).

Using the integration by parts formula:

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

Substitute the values:

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

Now, integrate ( -\frac{1}{2} \sin^2(x) ):

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

This integral can be solved using the half-angle identity ( \sin^2(x) = \frac{1 - \cos(2x)}{2} ):

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

[ = \frac{1}{4} \left( x - \frac{1}{2} \sin(2x) \right) + C ]

Putting it all together:

[ \int x \sin(x) \cos(x) , dx = -\frac{1}{2}x \sin^2(x) - \frac{1}{4} \left( x - \frac{1}{2} \sin(2x) \right) + C ]

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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.

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.

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.

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.

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