If `A=int_(0)^(pi)(sinx)/(sinx+cosx)dx, B=int_(0)^(pi)(sinx)/(sinx-cosx)dx`, then

Options

A. `A+B=0`
B. `A=B`
C. `A=B=pi//2`
D. `A=-B=pi`

Answer

To solve the problem, we need to evaluate the integrals \( A \) and \( B \) given by:

\[ A = \int_0^{\pi} \frac{\sin x}{\sin x + \cos x} \, dx \]
\[ B = \int_0^{\pi} \frac{\sin x}{\sin x - \cos x} \, dx \]

### Step 1: Evaluate \( A \)

We start with the integral \( A \):

\[ A = \int_0^{\pi} \frac{\sin x}{\sin x + \cos x} \, dx \]

Using the property of definite integrals, we can change the variable by letting \( x = \pi - t \). Then \( dx = -dt \), and the limits change from \( 0 \) to \( \pi \) to \( \pi \) to \( 0 \):

\[ A = \int_{\pi}^{0} \frac{\sin(\pi - t)}{\sin(\pi - t) + \cos(\pi - t)} (-dt) = \int_0^{\pi} \frac{\sin t}{\sin t - \cos t} \, dt \]

### Step 2: Evaluate \( B \)

Now, we have:

\[ B = \int_0^{\pi} \frac{\sin x}{\sin x - \cos x} \, dx \]

### Step 3: Relate \( A \) and \( B \)

From the previous steps, we have:

\[ A = \int_0^{\pi} \frac{\sin x}{\sin x + \cos x} \, dx \]
\[ B = \int_0^{\pi} \frac{\sin x}{\sin x - \cos x} \, dx \]

Now, we can see that:

\[ A + B = \int_0^{\pi} \left( \frac{\sin x}{\sin x + \cos x} + \frac{\sin x}{\sin x - \cos x} \right) dx \]

### Step 4: Combine the fractions

Combining the fractions gives:

\[ A + B = \int_0^{\pi} \frac{\sin x (\sin x - \cos x) + \sin x (\sin x + \cos x)}{(\sin x + \cos x)(\sin x - \cos x)} \, dx \]

This simplifies to:

\[ A + B = \int_0^{\pi} \frac{2 \sin^2 x}{\sin^2 x - \cos^2 x} \, dx \]

### Step 5: Simplify further

Using the identity \( \sin^2 x - \cos^2 x = -\cos(2x) \):

\[ A + B = -2 \int_0^{\pi} \frac{\sin^2 x}{\cos(2x)} \, dx \]

### Step 6: Conclusion

From the symmetry and properties of the integrals, we can conclude that:

\[ A = B \]

Thus, the relation between \( A \) and \( B \) is:

\[ A = B \]