If `int(cosx-sinx)dx=sqrt(2)sin(x+alpha)+c, " then " alpha=`

Options

A. `pi/3`
B. `-pi/3`
C. `pi/4`
D. `-pi/4`

Answer

To solve the problem, we need to find the value of \( \alpha \) in the equation:

\[ \int (\cos x - \sin x) \, dx = \sqrt{2} \sin(x + \alpha) + C \]

### Step 1: Integrate the left-hand side
We start by integrating \( \cos x - \sin x \):

\[ \int (\cos x - \sin x) \, dx = \int \cos x \, dx - \int \sin x \, dx \]

Using the known integrals:

- \( \int \cos x \, dx = \sin x \)
- \( \int \sin x \, dx = -\cos x \)

Thus, we have:

\[ \int (\cos x - \sin x) \, dx = \sin x + \cos x + C \]

### Step 2: Rewrite the result
Now we need to express \( \sin x + \cos x \) in a form that resembles \( \sqrt{2} \sin(x + \alpha) \). We can factor out \( \sqrt{2} \):

\[ \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) \]

### Step 3: Identify the angle
We know that:

\[ \frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \]

Thus, we can rewrite our expression as:

\[ \sin x + \cos x = \sqrt{2} \left( \sin\left(\frac{\pi}{4}\right) \sin x + \cos\left(\frac{\pi}{4}\right) \cos x \right) \]

Using the sine addition formula:

\[ \sin A \cos B + \cos A \sin B = \sin(A + B) \]

we can write:

\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]

### Step 4: Final equation
Now, substituting this back into our integral result, we get:

\[ \sin x + \cos x + C = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) + C \]

### Step 5: Compare with the given equation
From the original equation:

\[ \sqrt{2} \sin(x + \alpha) + C \]

we can see that:

\[ \alpha = \frac{\pi}{4} \]

### Conclusion
Thus, the value of \( \alpha \) is:

\[ \alpha = \frac{\pi}{4} \]