Evaluating integrals dependent on a parameter:
Differentiate I with respect to the parameter within the sign an integrals taking variable of the integrand as constant. Now evaluate the integral so obtained as a function of the parameter then integrate then result of get I. Constant of integration can be computed by giving some arbitrary values to the parameter and the corresponding value of I.
The value`int_(0)^(pi//2)log(sin^(2)theta+k^(2)cos^(2)theta)d theta`, where `kge0,` is

Options

A. `pi log (1+k)+pilog2`
B. `pilog(1+k)`
C. `pi log (1+k)-pi log 2`
D. `log (1+k)-log 2`

Explanation

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin^2 \theta + k^2 \cos^2 \theta) \, d\theta \]

where \( k \geq 0 \), we will follow the steps outlined in the problem statement.

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

We start by differentiating \( I \) with respect to \( k \):

\[ \frac{dI}{dk} = \int_{0}^{\frac{\pi}{2}} \frac{\partial}{\partial k} \log(\sin^2 \theta + k^2 \cos^2 \theta) \, d\theta \]

Using the chain rule, we have:

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

Thus,

\[ \frac{dI}{dk} = \int_{0}^{\frac{\pi}{2}} \frac{2k \cos^2 \theta}{\sin^2 \theta + k^2 \cos^2 \theta} \, d\theta \]

### Step 2: Evaluate the integral obtained

Now we need to evaluate the integral:

\[ \frac{dI}{dk} = 2k \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 \theta}{\sin^2 \theta + k^2 \cos^2 \theta} \, d\theta \]

To simplify this 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 from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) which corresponds to \( t = 0 \) to \( t = \infty \):

\[ \frac{dI}{dk} = 2k \int_{0}^{\infty} \frac{\frac{1}{1+t^2}}{\frac{t^2}{1+t^2} + k^2 \cdot \frac{1}{1+t^2}} \cdot \frac{dt}{1+t^2} \]

This simplifies to:

\[ \frac{dI}{dk} = 2k \int_{0}^{\infty} \frac{1}{t^2 + k^2} \, dt \]

The integral \( \int_{0}^{\infty} \frac{1}{t^2 + k^2} \, dt \) is known to be:

\[ \int_{0}^{\infty} \frac{1}{t^2 + k^2} \, dt = \frac{\pi}{2k} \]

Thus,

\[ \frac{dI}{dk} = 2k \cdot \frac{\pi}{2k} = \pi \]

### Step 3: Integrate \( \frac{dI}{dk} \)

Now we integrate \( \frac{dI}{dk} \):

\[ I = \int \pi \, dk = \pi k + C \]

where \( C \) is the constant of integration.

### Step 4: Determine the constant of integration

To find \( C \), we can evaluate \( I \) at \( k = 0 \):

\[ I(0) = \int_{0}^{\frac{\pi}{2}} \log(\sin^2 \theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} 2 \log(\sin \theta) \, d\theta \]

Using the known result \( \int_{0}^{\frac{\pi}{2}} \log(\sin \theta) \, d\theta = -\frac{\pi}{2} \):

\[ I(0) = 2 \left(-\frac{\pi}{2}\right) = -\pi \]

Thus,

\[ -\pi = \pi \cdot 0 + C \implies C = -\pi \]

### Final Result

Therefore, the integral \( I \) can be expressed as:

\[ I = \pi k - \pi = \pi(k - 1) \]

### Conclusion

The value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} \log(\sin^2 \theta + k^2 \cos^2 \theta) \, d\theta = \pi \log(1 + k) \]