`int_(a)^(b) (sin (x-a) -cos (x-a))/(sin (b-x)- cos (b-x)) dx = int_(a)^(b) (sin (b-x) -cos (b-x))/(sin (x-a) -cos (x-a))dx`
True or False?

Answer

To determine whether the given statement is true or false, we will evaluate both sides of the equation:

\[ \int_{a}^{b} \frac{\sin(x-a) - \cos(x-a)}{\sin(b-x) - \cos(b-x)} \, dx = \int_{a}^{b} \frac{\sin(b-x) - \cos(b-x)}{\sin(x-a) - \cos(x-a)} \, dx \]

### Step 1: Consider the left-hand side

Let:
\[ I = \int_{a}^{b} \frac{\sin(x-a) - \cos(x-a)}{\sin(b-x) - \cos(b-x)} \, dx \]

### Step 2: Apply the property of definite integrals

Using the property of definite integrals known as King's Rule, we can transform the integral. According to King's Rule:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]

Here, we set \( f(x) = \frac{\sin(x-a) - \cos(x-a)}{\sin(b-x) - \cos(b-x)} \).

### Step 3: Substitute \( x \) with \( a + b - x \)

Now, we will substitute \( x \) with \( a + b - x \) in the integral:

\[ I = \int_{a}^{b} \frac{\sin((a+b-x)-a) - \cos((a+b-x)-a)}{\sin(b-(a+b-x)) - \cos(b-(a+b-x))} \, dx \]

This simplifies to:

\[ I = \int_{a}^{b} \frac{\sin(b-x) - \cos(b-x)}{\sin(x-a) - \cos(x-a)} \, dx \]

### Step 4: Compare with the right-hand side

Now we have:

\[ I = \int_{a}^{b} \frac{\sin(b-x) - \cos(b-x)}{\sin(x-a) - \cos(x-a)} \, dx \]

This shows that the left-hand side is equal to the right-hand side.

### Conclusion

Since both sides of the equation are equal, we conclude that the statement is **True**.