Consider the integrals
`A = int_(0)^(pi) (sinxdx)/(sinx + cos x)` and `B = int_(0)^(pi) (sinxdx)/(sinx - cos x)`
Which one of the following is correct ?

Options

A. `A = 2B`
B. `B = 2A`
C. `A= B`
D. `A = 3B`

Explanation

To solve the integrals \( A \) and \( B \) given by:

\[ A = \int_{0}^{\pi} \frac{\sin x \, dx}{\sin x + \cos x} \]

\[ B = \int_{0}^{\pi} \frac{\sin x \, dx}{\sin x - \cos x} \]

we will analyze both integrals step by step.

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

The integral \( A \) is given as:

\[ A = \int_{0}^{\pi} \frac{\sin x \, dx}{\sin x + \cos x} \]

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

Now, we will evaluate \( B \):

\[ B = \int_{0}^{\pi} \frac{\sin x \, dx}{\sin x - \cos x} \]

To simplify \( B \), we can use the property of definite integrals, specifically the substitution \( x = \pi - t \).

### Step 3: Apply Substitution to \( B \)

Let \( x = \pi - t \). Then, \( dx = -dt \) and the limits change as follows:
- When \( x = 0 \), \( t = \pi \)
- When \( x = \pi \), \( t = 0 \)

Thus, we rewrite \( B \):

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

### Step 4: Compare \( A \) and \( B \)

Now we observe that:

\[ B = \int_{0}^{\pi} \frac{\sin t \, dt}{\sin t + \cos t} = A \]

### Conclusion

Thus, we find that:

\[ A = B \]

### Final Answer

The correct relationship is:

\[ A = B \]