2021-11-19

Evaluate the integral.
$\int \frac{\mathrm{cos}\left(1-\mathrm{ln}\left(y\right)\right)}{y}dy$

huckelig75

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

Provere

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