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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.
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