The value of `f(k)=int_(0)^(pi//2) log (sin ^(2) theta +k^(2) cos^(2) theta) d theta` is equal to

Options

A. `pi log (1+k)- pi log ^(2)`
B. `pi log 2 - log (1+k)`
C. `log(1+k) - pi log 2`
D. None of these

Answer

To solve the problem, we need to evaluate the integral

\[ f(k) = \int_0^{\frac{\pi}{2}} \log(\sin^2 \theta + k^2 \cos^2 \theta) \, d\theta. \]

### Step 1: Use the property of definite integrals

We can use the property of definite integrals that states:

\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx. \]

Applying this property, we have:

\[ f(k) = \int_0^{\frac{\pi}{2}} \log(\cos^2 \theta + k^2 \sin^2 \theta) \, d\theta. \]

### Step 2: Differentiate with respect to \( k \)

Now, we differentiate \( f(k) \) with respect to \( k \):

\[ \frac{df}{dk} = \int_0^{\frac{\pi}{2}} \frac{d}{dk} \log(\sin^2 \theta + k^2 \cos^2 \theta) \, d\theta. \]

Using the chain rule, we get:

\[ \frac{d}{dk} \log(\sin^2 \theta + k^2 \cos^2 \theta) = \frac{2k \cos^2 \theta}{\sin^2 \theta + k^2 \cos^2 \theta}. \]

Thus,

\[ \frac{df}{dk} = \int_0^{\frac{\pi}{2}} \frac{2k \cos^2 \theta}{\sin^2 \theta + k^2 \cos^2 \theta} \, d\theta. \]

### Step 3: Change of variable

To simplify the integral, we can use the substitution \( t = \tan \theta \), which gives \( \sin^2 \theta = \frac{t^2}{1+t^2} \) and \( \cos^2 \theta = \frac{1}{1+t^2} \). The differential \( d\theta \) becomes \( \frac{dt}{1+t^2} \).

The limits change as follows: when \( \theta = 0 \), \( t = 0 \) and when \( \theta = \frac{\pi}{2} \), \( t \to \infty \).

Thus, we have:

\[ \frac{df}{dk} = \int_0^{\infty} \frac{2k \cdot \frac{1}{1+t^2}}{\frac{t^2}{1+t^2} + k^2 \cdot \frac{1}{1+t^2}} \cdot \frac{dt}{1+t^2}. \]

This simplifies to:

\[ \frac{df}{dk} = 2k \int_0^{\infty} \frac{dt}{(1+t^2)(t^2 + k^2)}. \]

### Step 4: Solve the integral

The integral

\[ \int_0^{\infty} \frac{dt}{(1+t^2)(t^2 + k^2)} \]

can be evaluated using partial fractions or known results. The result is:

\[ \int_0^{\infty} \frac{dt}{(1+t^2)(t^2 + k^2)} = \frac{\pi}{2(1+k)}. \]

Thus,

\[ \frac{df}{dk} = 2k \cdot \frac{\pi}{2(1+k)} = \frac{k\pi}{1+k}. \]

### Step 5: Integrate to find \( f(k) \)

Now we integrate \( \frac{df}{dk} \) with respect to \( k \):

\[ f(k) = \int \frac{k\pi}{1+k} \, dk. \]

Using the substitution \( u = 1 + k \), we have \( du = dk \) and \( k = u - 1 \):

\[ f(k) = \int \frac{(u-1)\pi}{u} \, du = \pi \int (1 - \frac{1}{u}) \, du = \pi \left( u - \log u \right) + C. \]

Substituting back \( u = 1 + k \):

\[ f(k) = \pi \left( (1 + k) - \log(1 + k) \right) + C. \]

### Step 6: Determine the constant \( C \)

To find \( C \), we can evaluate \( f(0) \):

\[ f(0) = \int_0^{\frac{\pi}{2}} \log(\sin^2 \theta) \, d\theta = -\frac{\pi}{2} \log(2). \]

Setting \( k = 0 \):

\[ f(0) = \pi (1 - \log(1)) + C = \pi + C. \]

Thus,

\[ -\frac{\pi}{2} \log(2) = \pi + C \implies C = -\frac{\pi}{2} \log(2) - \pi. \]

### Final Result

The final expression for \( f(k) \) is:

\[ f(k) = \pi \left( (1 + k) - \log(1 + k) \right) - \frac{\pi}{2} \log(2) - \pi. \]

This simplifies to:

\[ f(k) = \pi \left( k - \log(1 + k) - \frac{1}{2} \log(2) \right). \]