`If|a|lt 1,` show that `int _(0)^(pi)(log(1+a cos x ))/( cos x)dx =pi sin^(-1) a `

Answer

To solve the integral

\[ I(a) = \int_0^\pi \frac{\log(1 + a \cos x)}{\cos x} \, dx \]

where \(|a| < 1\), we will differentiate \(I(a)\) with respect to \(a\) and then integrate back to find \(I(a)\).

### Step 1: Differentiate \(I(a)\) with respect to \(a\)

Using Leibniz's rule for differentiation under the integral sign, we have:

\[ I'(a) = \int_0^\pi \frac{\partial}{\partial a} \left( \frac{\log(1 + a \cos x)}{\cos x} \right) dx \]

Calculating the partial derivative:

\[ \frac{\partial}{\partial a} \left( \log(1 + a \cos x) \right) = \frac{\cos x}{1 + a \cos x} \]

Thus, we can write:

\[ I'(a) = \int_0^\pi \frac{\cos x}{1 + a \cos x} \cdot \frac{1}{\cos x} \, dx = \int_0^\pi \frac{1}{1 + a \cos x} \, dx \]

### Step 2: Evaluate the integral \( \int_0^\pi \frac{1}{1 + a \cos x} \, dx \)

This integral can be evaluated using the known result:

\[ \int_0^\pi \frac{1}{1 + a \cos x} \, dx = \frac{\pi}{\sqrt{1 - a^2}} \quad \text{for } |a| < 1 \]

Thus, we have:

\[ I'(a) = \frac{\pi}{\sqrt{1 - a^2}} \]

### Step 3: Integrate \(I'(a)\) to find \(I(a)\)

Now, we integrate \(I'(a)\) with respect to \(a\):

\[ I(a) = \int I'(a) \, da = \int \frac{\pi}{\sqrt{1 - a^2}} \, da \]

The integral of \(\frac{1}{\sqrt{1 - a^2}}\) is:

\[ \int \frac{1}{\sqrt{1 - a^2}} \, da = \sin^{-1}(a) + C \]

Thus, we have:

\[ I(a) = \pi \sin^{-1}(a) + C \]

### Step 4: Determine the constant \(C\)

To find the constant \(C\), we can evaluate \(I(0)\):

\[ I(0) = \int_0^\pi \frac{\log(1 + 0 \cdot \cos x)}{\cos x} \, dx = \int_0^\pi \frac{\log(1)}{\cos x} \, dx = 0 \]

Now substituting \(a = 0\) into our expression for \(I(a)\):

\[ I(0) = \pi \sin^{-1}(0) + C = 0 + C \implies C = 0 \]

### Final Result

Thus, we conclude that:

\[ I(a) = \pi \sin^{-1}(a) \]

Therefore, we have shown that

\[ \int_0^\pi \frac{\log(1 + a \cos x)}{\cos x} \, dx = \pi \sin^{-1}(a) \]