# What is y if dy/dx=sin(lnx)/x?

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Given dy/dx = sin(lnx)/x

We need to find y.

Then we will find the integral of dy/dx.

Let u = ln x ==> du = 1/x dx ==> xdu = dx

==> dy/dx = sinu / x

==> dy = sin(u) /x * dx

==> Int dy = Int sin(u)/x * x du

==> y = Int sinu du

==> y= -cos u + C

Now we will substitute with u= ln x

**==> y= -cos ( lnx ) + C**

To determine the primitive y, we'll have to calculate the indefinite integral of dy.

We'll solve by replacing ln x by t:

ln x = t

We'll differentiate both sides:

dx/x = dt

We'll re-write the integral in the new variable t:

Int dy = Int sin(lnx) dx/x

Int sin(lnx) dx/x = Int sin t dt

Int sin t dt = - cos t + C

We'll replace t by ln x:

Int sin(lnx) dx/x = - cos (ln x) + C

**The primitive function is: y = - cos (ln x) + C.**