Let `f(x) = ln ((1-sinx)/(1+sinx))`, then show that `int_(a)^(b) f(x)dx = int_(b)^(a) ln((1+sinx)/(1-sinx))dx`.

Answer

To prove that

\[ \int_{a}^{b} f(x) \, dx = \int_{b}^{a} \ln\left(\frac{1+\sin x}{1-\sin x}\right) \, dx \]

where \( f(x) = \ln\left(\frac{1 - \sin x}{1 + \sin x}\right) \), we will follow these steps:

### Step 1: Write the integral of \( f(x) \)

We start with the left-hand side:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} \ln\left(\frac{1 - \sin x}{1 + \sin x}\right) \, dx \]

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

Using the property of definite integrals, we know that:

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

Thus, we can write:

\[ \int_{a}^{b} \ln\left(\frac{1 - \sin x}{1 + \sin x}\right) \, dx = -\int_{b}^{a} \ln\left(\frac{1 - \sin x}{1 + \sin x}\right) \, dx \]

### Step 3: Change the limits of integration

When we change the limits of integration, we can express this as:

\[ -\int_{b}^{a} \ln\left(\frac{1 - \sin x}{1 + \sin x}\right) \, dx = \int_{a}^{b} \ln\left(\frac{1 + \sin x}{1 - \sin x}\right) \, dx \]

### Step 4: Apply the logarithmic property

Using the property of logarithms, we have:

\[ -\ln\left(\frac{1 - \sin x}{1 + \sin x}\right) = \ln\left(\frac{1 + \sin x}{1 - \sin x}\right) \]

### Step 5: Combine the results

Thus, we can conclude:

\[ \int_{a}^{b} \ln\left(\frac{1 - \sin x}{1 + \sin x}\right) \, dx = \int_{b}^{a} \ln\left(\frac{1 + \sin x}{1 - \sin x}\right) \, dx \]

This shows that:

\[ \int_{a}^{b} f(x) \, dx = \int_{b}^{a} \ln\left(\frac{1 + \sin x}{1 - \sin x}\right) \, dx \]

### Conclusion

Thus, we have proved the required identity.

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