`int(tanx)/((1-sinx))dx`

Explanation

To solve the integral \(\int \frac{\tan x}{1 - \sin x} \, dx\), we will follow a step-by-step approach.

### Step 1: Rewrite the Integral
We start by rewriting \(\tan x\) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we can rewrite the integral as:
\[ \int \frac{\tan x}{1 - \sin x} \, dx = \int \frac{\sin x}{\cos x (1 - \sin x)} \, dx \]

### Step 2: Simplify the Expression
Next, we can simplify the expression:
\[ \int \frac{\sin x}{\cos x (1 - \sin x)} \, dx = \int \frac{\sin x}{\cos^2 x (1 - \sin x)} \, dx \]

### Step 3: Use Substitution
To integrate this expression, we can use the substitution \(u = 1 - \sin x\). Then, we find \(du\):
\[ du = -\cos x \, dx \quad \Rightarrow \quad dx = -\frac{du}{\cos x} \]
Also, we need to express \(\sin x\) in terms of \(u\):
\[ \sin x = 1 - u \]
Now, substituting these into the integral gives:
\[ \int \frac{1 - u}{\cos^2 x \cdot u} \left(-\frac{du}{\cos x}\right) = -\int \frac{(1 - u)}{u \cos^3 x} \, du \]

### Step 4: Express \(\cos^2 x\) in terms of \(u\)
From the identity \(\sin^2 x + \cos^2 x = 1\):
\[ \cos^2 x = 1 - \sin^2 x = 1 - (1 - u)^2 = 1 - (1 - 2u + u^2) = 2u - u^2 \]
Thus, \(\cos^3 x = (2u - u^2)^{3/2}\).

### Step 5: Substitute Back into the Integral
Now we can substitute back into the integral:
\[ -\int \frac{(1 - u)}{u (2u - u^2)^{3/2}} \, du \]

### Step 6: Solve the Integral
This integral can be solved using standard integration techniques, such as partial fractions or further substitution.

### Step 7: Back Substitute
After integrating, we will substitute back \(u = 1 - \sin x\) to express the final answer in terms of \(x\).

### Final Answer
The final answer will be in the form of a function of \(x\) plus a constant of integration \(C\).