Evaluate the following integrals:
`int (1-sinx )/(x+cos x) dx`

Answer

To evaluate the integral

\[ \int \frac{1 - \sin x}{x + \cos x} \, dx, \]

we can follow these steps:

### Step 1: Identify the substitution
We notice that the numerator \(1 - \sin x\) is the derivative of the denominator \(x + \cos x\). Therefore, we can use the substitution:

\[ t = x + \cos x. \]

### Step 2: Differentiate the substitution
Now, we differentiate \(t\) with respect to \(x\):

\[ \frac{dt}{dx} = 1 - \sin x. \]

This implies:

\[ dt = (1 - \sin x) \, dx. \]

### Step 3: Rewrite the integral
We can now rewrite the integral in terms of \(t\):

\[ dx = \frac{dt}{1 - \sin x}. \]

Substituting \(t\) and \(dt\) into the integral gives:

\[ \int \frac{1 - \sin x}{x + \cos x} \, dx = \int \frac{1 - \sin x}{t} \cdot \frac{dt}{1 - \sin x} = \int \frac{dt}{t}. \]

### Step 4: Integrate
The integral \(\int \frac{dt}{t}\) is a standard integral:

\[ \int \frac{dt}{t} = \log |t| + C, \]

where \(C\) is the constant of integration.

### Step 5: Substitute back
Now we substitute back \(t = x + \cos x\):

\[ \log |x + \cos x| + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int \frac{1 - \sin x}{x + \cos x} \, dx = \log |x + \cos x| + C. \]

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