`int_(0)^(pi//2)log (sec x) dx=`

Options

A. `(-pi)/2log2`
B. `pi/2 log 2`
C. `(-pi)/4 log 2`
D. `pi/4log2`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \log(\sec x) \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We know that \( \sec x = \frac{1}{\cos x} \), so we can rewrite the logarithm:
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sec x) \, dx = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{\cos x}\right) \, dx = \int_{0}^{\frac{\pi}{2}} -\log(\cos x) \, dx \]
Thus, we have:
\[ I = -\int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \]

### Step 2: Use the property of integrals
Using the property of integrals, we can express \( I \) in terms of another integral:
\[ I = -\int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \]
Now, we can also express this integral using the substitution \( x = \frac{\pi}{2} - t \):
\[ \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \]
Thus, we have:
\[ I = -\int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]

### Step 3: Combine the integrals
Now we can add both expressions for \( I \):
\[ 2I = -\int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx - \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]
This simplifies to:
\[ 2I = -\int_{0}^{\frac{\pi}{2}} \left( \log(\sin x) + \log(\cos x) \right) \, dx \]
Using the property of logarithms, we can combine the logs:
\[ \log(\sin x) + \log(\cos x) = \log(\sin x \cos x) \]
Thus:
\[ 2I = -\int_{0}^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx \]

### Step 4: Simplify the integral
Using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \):
\[ 2I = -\int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2} \sin(2x)\right) \, dx \]
This can be split into two integrals:
\[ 2I = -\int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx - \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx \]
Calculating the first integral:
\[ -\int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx = -\frac{\pi}{2} \log\left(\frac{1}{2}\right) = \frac{\pi}{2} \log(2) \]

### Step 5: Change of variables for the second integral
For the second integral, let \( u = 2x \), then \( du = 2dx \) or \( dx = \frac{du}{2} \). The limits change from \( 0 \) to \( \pi \):
\[ -\int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx = -\frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du \]
We know that:
\[ \int_{0}^{\pi} \log(\sin u) \, du = -\pi \log(2) \]
Thus:
\[ -\frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du = -\frac{1}{2} (-\pi \log(2)) = \frac{\pi}{2} \log(2) \]

### Step 6: Combine results
Putting it all together:
\[ 2I = \frac{\pi}{2} \log(2) + \frac{\pi}{2} \log(2) = \pi \log(2) \]
Thus:
\[ I = \frac{\pi}{2} \log(2) \]

### Final Answer
\[ \int_{0}^{\frac{\pi}{2}} \log(\sec x) \, dx = \frac{\pi}{2} \log(2) \]