Evaluate the following integrals.
` int (sinx + cos x)/((sin x - cos x)^(3))dx `

Answer

To evaluate the integral

\[ I = \int \frac{\sin x + \cos x}{(\sin x - \cos x)^3} \, dx, \]

we will use the substitution method.

### Step 1: Choose a substitution
Let

\[ t = \sin x - \cos x. \]

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

\[ \frac{dt}{dx} = \cos x + \sin x \quad \Rightarrow \quad dt = (\cos x + \sin x) \, dx. \]

### Step 3: Rearranging for \(dx\)
From the above, we can express \(dx\) in terms of \(dt\):

\[ dx = \frac{dt}{\cos x + \sin x}. \]

### Step 4: Substitute in the integral
Now, we substitute \(t\) and \(dx\) back into the integral:

\[ I = \int \frac{\sin x + \cos x}{t^3} \cdot \frac{dt}{\cos x + \sin x}. \]

Notice that \(\sin x + \cos x\) in the numerator and \(\cos x + \sin x\) in the denominator cancel out:

\[ I = \int \frac{1}{t^3} \, dt. \]

### Step 5: Integrate
Now we can integrate:

\[ I = \int t^{-3} \, dt = \frac{t^{-2}}{-2} + C = -\frac{1}{2t^2} + C. \]

### Step 6: Substitute back for \(t\)
Now we substitute back \(t = \sin x - \cos x\):

\[ I = -\frac{1}{2(\sin x - \cos x)^2} + C. \]

Thus, the final answer is:

\[ I = -\frac{1}{2(\sin x - \cos x)^2} + C. \]

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