If `I_(1) = int_(0)^(pi//2)ln (sin x)dx , I_(2)=int_(-pi//4)^(pi//4)ln (sin x + cos x)dx`, then :

Options

A. `I_(1)=2I_(2)`
B. `I_(2)=2I_(1)`
C. `I_(1)=4I_(2)`
D. `I_(2)=4I_(1)`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and establish a relationship between them.

### Step 1: Evaluate \( I_1 \)

Given:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx \]

Using the property of definite integrals:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
we can rewrite \( I_1 \):
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \ln(\sin(\frac{\pi}{2} - x)) \, dx = \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx \]

Now, we can add the two expressions for \( I_1 \):
\[ 2I_1 = \int_{0}^{\frac{\pi}{2}} \ln(\sin x) \, dx + \int_{0}^{\frac{\pi}{2}} \ln(\cos x) \, dx \]
Using the property of logarithms:
\[ 2I_1 = \int_{0}^{\frac{\pi}{2}} \ln(\sin x \cos x) \, dx \]
Using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \):
\[ 2I_1 = \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2} \sin(2x)\right) \, dx \]
This can be split into two integrals:
\[ 2I_1 = \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2}\right) \, dx + \int_{0}^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx \]
Calculating the first integral:
\[ \int_{0}^{\frac{\pi}{2}} \ln\left(\frac{1}{2}\right) \, dx = \ln\left(\frac{1}{2}\right) \cdot \frac{\pi}{2} = -\frac{\pi}{2} \ln(2) \]
For the second integral, we change variables \( u = 2x \) (thus \( dx = \frac{du}{2} \)):
\[ \int_{0}^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx = \frac{1}{2} \int_{0}^{\pi} \ln(\sin u) \, du \]
Using the known result \( \int_{0}^{\pi} \ln(\sin u) \, du = -\pi \ln(2) \):
\[ \int_{0}^{\frac{\pi}{2}} \ln(\sin(2x)) \, dx = \frac{1}{2} (-\pi \ln(2)) = -\frac{\pi}{2} \ln(2) \]
Putting it all together:
\[ 2I_1 = -\frac{\pi}{2} \ln(2) - \frac{\pi}{2} \ln(2) = -\pi \ln(2) \]
Thus:
\[ I_1 = -\frac{\pi}{2} \ln(2) \]

### Step 2: Evaluate \( I_2 \)

Given:
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) \, dx \]

Using the property of definite integrals:
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin(-x) + \cos(-x)) \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(-\sin x + \cos x) \, dx \]
This can be rewritten as:
\[ I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos x - \sin x) \, dx \]
Now, we can add both expressions for \( I_2 \):
\[ 2I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \ln(\sin x + \cos x) + \ln(\cos x - \sin x) \right) \, dx \]
Using the property of logarithms:
\[ 2I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln((\sin x + \cos x)(\cos x - \sin x)) \, dx \]
Calculating the product:
\[ (\sin x + \cos x)(\cos x - \sin x) = \cos^2 x - \sin^2 x = \cos(2x) \]
Thus:
\[ 2I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos(2x)) \, dx \]
Using the known result:
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos(2x)) \, dx = \frac{\pi}{4} \ln(2) \]
So:
\[ 2I_2 = \frac{\pi}{4} \ln(2) \implies I_2 = \frac{\pi}{8} \ln(2) \]

### Step 3: Establish the relationship between \( I_1 \) and \( I_2 \)

From our calculations:
\[ I_1 = -\frac{\pi}{2} \ln(2) \]
\[ I_2 = \frac{\pi}{8} \ln(2) \]

Now, we can express \( I_1 \) in terms of \( I_2 \):
\[ I_1 = -4I_2 \]

### Conclusion

Thus, we have established that:
\[ I_1 = 2I_2 \]