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How do you evaluate the integral of #int (cos(lnx))/x dx#?

Answer 1

Charles Anctil

#intcos(lnx)/xdx=sin(lnx)+C#

A substitution solves this.

#u=lnx#

#du=dx/x#

Then, we have

#intcosudu=sinu+C#

Rewriting in terms of #x# yields

#intcos(lnx)/xdx=sin(lnx)+C#

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Answer 2

Isabella Ackerman

To evaluate the integral ∫(cos(lnx))/x dx, we can use the substitution method. Let u = ln(x), then du = (1/x)dx. This transforms the integral into ∫cos(u)du. Integrating cos(u) gives sin(u) + C. Substituting back u = ln(x), we get sin(ln(x)) + C as the final result.

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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.