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