Let `u=int_0^1("ln"(x+1))/(x^2+1)dxa n dv=int_0^(pi/2)ln(sin2x)dx ,t h e n` `u=-pi/2ln2` (b) `4u+v=0` `u+4v=0` (d) `u=pi/8ln2`

Options

A. `u=4v`
B. `4u+v=0`
C. `u+4v=0`
D. `2u+v=0`

Answer

To solve the problem, we need to evaluate the integrals \( u \) and \( v \) given by:

1. \( u = \int_0^1 \frac{\ln(x+1)}{x^2+1} \, dx \)
2. \( v = \int_0^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx \)

We will then check which of the provided options are correct based on the values of \( u \) and \( v \).

### Step 1: Evaluate \( u \)

We start by evaluating \( u \):

\[ u = \int_0^1 \frac{\ln(x+1)}{x^2+1} \, dx \]

To solve this integral, we can use the substitution \( x = \tan(\theta) \), which gives us \( dx = \sec^2(\theta) \, d\theta \). The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x = 1 \), \( \theta = \frac{\pi}{4} \)

Thus, we have:

\[ u = \int_0^{\frac{\pi}{4}} \frac{\ln(1 + \tan(\theta))}{1 + \tan^2(\theta)} \sec^2(\theta) \, d\theta \]

Since \( 1 + \tan^2(\theta) = \sec^2(\theta) \), we can simplify:

\[ u = \int_0^{\frac{\pi}{4}} \ln(1 + \tan(\theta)) \, d\theta \]

Next, we can express \( \ln(1 + \tan(\theta)) \) in terms of sine and cosine:

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

Thus, we can split the integral:

\[ u = \int_0^{\frac{\pi}{4}} \ln(\sin(\theta) + \cos(\theta)) \, d\theta - \int_0^{\frac{\pi}{4}} \ln(\cos(\theta)) \, d\theta \]

The second integral can be calculated using known results:

\[ \int_0^{\frac{\pi}{4}} \ln(\cos(\theta)) \, d\theta = -\frac{\pi}{4} \ln(2) \]

Now, we need to evaluate the first integral. Using properties of logarithms and symmetry, we can find that:

\[ \int_0^{\frac{\pi}{4}} \ln(\sin(\theta) + \cos(\theta)) \, d\theta = \frac{\pi}{4} \ln(2) \]

Putting it all together, we find:

\[ u = \frac{\pi}{4} \ln(2) + \frac{\pi}{4} \ln(2) = \frac{\pi}{8} \ln(2) \]

### Step 2: Evaluate \( v \)

Now we evaluate \( v \):

\[ v = \int_0^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx \]

Using the substitution \( 2x = t \), we have \( dx = \frac{1}{2} dt \). The limits change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = \frac{\pi}{2} \), \( t = \pi \)

Thus, we have:

\[ v = \frac{1}{2} \int_0^{\pi} \ln(\sin(t)) \, dt \]

Using the known result:

\[ \int_0^{\pi} \ln(\sin(t)) \, dt = -\pi \ln(2) \]

We find:

\[ v = \frac{1}{2} \cdot (-\pi \ln(2)) = -\frac{\pi}{2} \ln(2) \]

### Step 3: Check the Options

Now we have:

- \( u = \frac{\pi}{8} \ln(2) \)
- \( v = -\frac{\pi}{2} \ln(2) \)

Now we check the options:

(a) \( u = -4v \)

Calculating \( -4v \):

\[ -4v = -4 \left(-\frac{\pi}{2} \ln(2)\right) = 2\pi \ln(2) \quad \text{(not equal to } u\text{)} \]

(b) \( 4u + v = 0 \)

Calculating \( 4u + v \):

\[ 4u = 4 \left(\frac{\pi}{8} \ln(2)\right) = \frac{\pi}{2} \ln(2) \]
\[ 4u + v = \frac{\pi}{2} \ln(2) - \frac{\pi}{2} \ln(2) = 0 \quad \text{(this is correct)} \]

(c) \( u + 4v = 0 \)

Calculating \( u + 4v \):

\[ 4v = 4 \left(-\frac{\pi}{2} \ln(2)\right) = -2\pi \ln(2) \]
\[ u + 4v = \frac{\pi}{8} \ln(2) - 2\pi \ln(2) \quad \text{(not equal to } 0\text{)} \]

(d) \( u = \frac{\pi}{8} \ln(2) \)

This is true as we calculated.

### Conclusion

The correct options are:

- (b) \( 4u + v = 0 \)
- (d) \( u = \frac{\pi}{8} \ln(2) \)