`I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx`,
`I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1)dx`. Then

Options

A. `I_(2)=I_(3)=I_(4)=0,I_(1)!=0`
B. `I_(1)=I_(2)=I_(3)=0,I_(4)!=0`
C. `I_(1)=I_(3)=I_(4)=0,I_(2)!=0`
D. `I_(1)=I_(2)=I_(3)=0,I_(4)!=0`

Explanation

To solve the problem, we need to evaluate the integrals \( I_1, I_2, I_3, \) and \( I_4 \) and find the relationships between them.

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

We start with the integral:
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \]

Using the substitution \( x = \frac{\pi}{2} - t \), we have:
\[ \sin x = \cos t \quad \text{and} \quad \cos x = \sin t \]
Thus,
\[ I_1 = \int_{\frac{\pi}{2}}^0 \frac{\cos t - \sin t}{1 + \cos t \sin t} (-dt) = \int_0^{\frac{\pi}{2}} \frac{\cos t - \sin t}{1 + \cos t \sin t} \, dt \]

Now, we can rewrite \( I_1 \):
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx \]
Adding these two expressions for \( I_1 \):
\[ 2I_1 = \int_0^{\frac{\pi}{2}} \frac{(\sin x - \cos x) + (\cos x - \sin x)}{1 + \sin x \cos x} \, dx = 0 \]
Thus, we find:
\[ I_1 = 0 \]

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

Next, we evaluate:
\[ I_2 = \int_0^{2\pi} \cos^6 x \, dx \]

Using the periodicity of cosine, we can write:
\[ I_2 = 2 \int_0^{\pi} \cos^6 x \, dx \]

Using the reduction formula or known integrals, we find:
\[ \int_0^{\pi} \cos^6 x \, dx = \frac{5\pi}{16} \]
Thus,
\[ I_2 = 2 \cdot \frac{5\pi}{16} = \frac{5\pi}{8} \]

### Step 3: Evaluate \( I_3 \)

Now, we evaluate:
\[ I_3 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx \]

Since \( \sin^3 x \) is an odd function, we have:
\[ I_3 = 0 \]

### Step 4: Evaluate \( I_4 \)

Finally, we evaluate:
\[ I_4 = \int_0^1 \ln\left(\frac{1}{x} - 1\right) \, dx \]
This can be rewritten as:
\[ I_4 = \int_0^1 \ln\left(\frac{1-x}{x}\right) \, dx = \int_0^1 \ln(1-x) \, dx - \int_0^1 \ln(x) \, dx \]

Both integrals can be evaluated:
\[ \int_0^1 \ln(1-x) \, dx = -1 \quad \text{and} \quad \int_0^1 \ln(x) \, dx = -1 \]
Thus,
\[ I_4 = -1 - (-1) = 0 \]

### Summary of Results

Now we summarize our results:
- \( I_1 = 0 \)
- \( I_2 = \frac{5\pi}{8} \)
- \( I_3 = 0 \)
- \( I_4 = 0 \)

### Conclusion

From our evaluations, we conclude:
- \( I_1 = I_3 = I_4 = 0 \)
- \( I_2 \neq 0 \)