We are given the equation $\int \frac{1}{tan(y)} dy = \int x dx$ and asked to solve it.

1. Problem Description

We are given the equation $\int \frac{1}{tan(y)} dy = \int x dx$ and asked to solve it.

2. Solution Steps

First, we rewrite $\frac{1}{tan(y)}$ as $cot(y)$. Thus, the equation becomes:

$\int cot(y) dy = \int x dx$

We know that $\int cot(y) dy = ln|sin(y)| + C_1$ and $\int x dx = \frac{x^2}{2} + C_2$.

Therefore, $ln|sin(y)| + C_1 = \frac{x^2}{2} + C_2$.

Combining the constants, we get $ln|sin(y)| = \frac{x^2}{2} + C$, where $C = C_2 - C_1$.

Taking the exponential of both sides, we have $|sin(y)| = e^{\frac{x^2}{2} + C} = e^{\frac{x^2}{2}}e^C$.

We can write $e^C$ as another constant $A$, so $|sin(y)| = Ae^{\frac{x^2}{2}}$.

Then, $sin(y) = \pm Ae^{\frac{x^2}{2}}$. Let $B = \pm A$, so $sin(y) = Be^{\frac{x^2}{2}}$.

Thus, $y = arcsin(Be^{\frac{x^2}{2}})$.

$\int cot(y)dy = ln|sin(y)| + C_1$

$\int xdx = \frac{x^2}{2} + C_2$

$ln|sin(y)| = \frac{x^2}{2} + C$, where $C=C_2 - C_1$

$|sin(y)| = e^{\frac{x^2}{2} + C} = e^{\frac{x^2}{2}}e^C$

$|sin(y)| = A e^{\frac{x^2}{2}}$, where $A = e^C$

$sin(y) = \pm A e^{\frac{x^2}{2}}$

$y = arcsin(\pm A e^{\frac{x^2}{2}})$

3. Final Answer

$y = arcsin(Ae^{\frac{x^2}{2}})$, where A is a constant.