# How do you integrate #sin( ln x )#?

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To integrate ( \sin(\ln x) ), use integration by parts, which is defined as ( \int u , dv = uv - \int v , du ). Choose ( u ) and ( dv ) such that ( du ) and ( v ) can be easily determined. For ( \int \sin(\ln x) , dx ), let:

- ( u = \sin(\ln x) ) which implies ( du = \frac{\cos(\ln x)}{x} , dx )
- ( dv = dx ) which implies ( v = x )

Then, according to the integration by parts formula:

[ \int \sin(\ln x) , dx = x \sin(\ln x) - \int x \cdot \frac{\cos(\ln x)}{x} , dx ]

[ = x \sin(\ln x) - \int \cos(\ln x) , dx ]

To integrate ( \cos(\ln x) ), apply integration by parts again with:

- ( u = \cos(\ln x) ) which implies ( du = -\frac{\sin(\ln x)}{x} , dx )
- ( dv = dx ) which implies ( v = x )

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

[ = x \cos(\ln x) + \int \sin(\ln x) , dx ]

Let ( I = \int \sin(\ln x) , dx ), then we have:

[ I = x \sin(\ln x) - (x \cos(\ln x) + I) ]

Rearranging for ( I ):

[ 2I = x \sin(\ln x) - x \cos(\ln x) ]

[ I = \frac{1}{2} x (\sin(\ln x) - \cos(\ln x)) + C ]

Thus, the integral of ( \sin(\ln x) ) with respect to ( x ) is:

[ \frac{1}{2} x (\sin(\ln x) - \cos(\ln x)) + 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.

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