If `int_(0)^((pi)/2)log(cosx)dx=-(pi)/2log2`, then `int_(0)^((pi)/2)log(cosecx)dx=`

Options

A. `(pi)/2-(pi)/2log2`
B. `-(pi)/2log2`
C. `(pi)/2log2`
D. `(pi)/2+(pi)/2log2`

Answer

To solve the problem, we need to find the value of the integral \( I_2 = \int_0^{\frac{\pi}{2}} \log(\csc x) \, dx \) given that \( I_1 = \int_0^{\frac{\pi}{2}} \log(\cos x) \, dx = -\frac{\pi}{2} \log 2 \).

### Step-by-step Solution:

1. **Define the Integrals**:
Let
\[ I_1 = \int_0^{\frac{\pi}{2}} \log(\cos x) \, dx \]
and
\[ I_2 = \int_0^{\frac{\pi}{2}} \log(\csc x) \, dx. \]

2. **Use the Identity for Cosecant**:
Recall that
\[ \csc x = \frac{1}{\sin x}. \]
Therefore, we can rewrite \( I_2 \) as:
\[ I_2 = \int_0^{\frac{\pi}{2}} \log(\csc x) \, dx = \int_0^{\frac{\pi}{2}} \log\left(\frac{1}{\sin x}\right) \, dx. \]

3. **Simplify the Integral**:
Using the property of logarithms, we have:
\[ \log\left(\frac{1}{\sin x}\right) = -\log(\sin x). \]
Thus, we can express \( I_2 \) as:
\[ I_2 = -\int_0^{\frac{\pi}{2}} \log(\sin x) \, dx. \]

4. **Relate \( I_1 \) and \( I_2 \)**:
We know from the properties of definite integrals that:
\[ \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \log(\cos x) \, dx. \]
Therefore, we can write:
\[ \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx = I_1. \]

5. **Substitute Back**:
Now substituting back into the expression for \( I_2 \):
\[ I_2 = -I_1. \]

6. **Substitute the Value of \( I_1 \)**:
Given that \( I_1 = -\frac{\pi}{2} \log 2 \), we substitute this value into the equation for \( I_2 \):
\[ I_2 = -\left(-\frac{\pi}{2} \log 2\right) = \frac{\pi}{2} \log 2. \]

### Final Answer:
Thus, the value of \( \int_0^{\frac{\pi}{2}} \log(\csc x) \, dx \) is:
\[ \int_0^{\frac{\pi}{2}} \log(\csc x) \, dx = \frac{\pi}{2} \log 2. \]