Evaluate the integral. ∫cos⁡(1−ln⁡(y))ydy

Step 1
We have to find the integrals:
${\displaystyle \int \frac{\cos (1-\ln \left(y\right))}{y}dy}$
We will find this integrals by substitution method
Let ${\displaystyle t=1-\ln \left(y\right)}$
Differentiating both sides with respect to y, we get
${\displaystyle t=1-\ln \left(y\right)}$
${\displaystyle dt=0-\frac{1}{y}dy}$
${\displaystyle -dt=\frac{dy}{y}}$
Step 2
Now finding integrals putting above value,
${\displaystyle \int \frac{\cos (1-\ln \left(y\right))}{y}dy=\int \cos (1-\ln \left(y\right))\frac{dy}{y}}$
${\displaystyle =\int \cos tdt}$
${\displaystyle =\sin t+c}$.
Since integration of cosine function is sine.
Now putting ${\displaystyle t=1-\ln \left(y\right)}$, we get
${\displaystyle =\sin (1-\ln \left(y\right))+c}$.
Hence, integrals of the given expression is ${\displaystyle \sin (1-\ln \left(y\right))+c}$.

Step 1: Use Integration by Substitution.
Let ${\displaystyle u=1-\ln y,du=-\frac{1}{y}dy}$
Step 2: Using u and du above, rewrite ${\displaystyle \int \frac{\cos (1-\ln \left(y\right))}{y}dy}$.
${\displaystyle \int -\cos udu}$
Step 3: Use Trigonometric Integration: the integral of ${\displaystyle \cos uis\sin u}$.
${\displaystyle -\sin u}$
Step 4: Substitute ${\displaystyle u=1-\ln y}$ back into the original integral.
${\displaystyle -\sin (1-\ln y)}$
Step 5: Add constant.
${\displaystyle -\sin (1-\ln y)+C}$