If `I_(1)=int_(0)^(2pi)sin^(3)xdx and I_(2)=int_(0)^(1)ln((1)/(x)-1)dx`, then

Options

A. `I_(1)+I_(2) gt 0`
B. `I+I_(2) lt0`
C. `I_(1)lt I_(2)`
D. `I_(1)=I_(2)`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and then compare their values.

### Step 1: Evaluate \( I_1 = \int_0^{2\pi} \sin^3 x \, dx \)

We can use the property of integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(b + a - x) \, dx \]
For \( I_1 \), we can apply this property:
\[ I_1 = \int_0^{2\pi} \sin^3 x \, dx = \int_0^{2\pi} \sin^3(2\pi - x) \, dx \]
Since \( \sin(2\pi - x) = -\sin x \), we have:
\[ I_1 = \int_0^{2\pi} (-\sin x)^3 \, dx = -\int_0^{2\pi} \sin^3 x \, dx \]
This gives us:
\[ I_1 = -I_1 \]
Adding \( I_1 \) to both sides:
\[ 2I_1 = 0 \implies I_1 = 0 \]

### Step 2: Evaluate \( I_2 = \int_0^1 \ln\left(\frac{1}{x} - 1\right) \, dx \)

We can rewrite the logarithmic term:
\[ I_2 = \int_0^1 \ln\left(\frac{1 - x}{x}\right) \, dx \]
Using the properties of logarithms:
\[ I_2 = \int_0^1 \left(\ln(1 - x) - \ln x\right) \, dx \]
This can be split into two integrals:
\[ I_2 = \int_0^1 \ln(1 - x) \, dx - \int_0^1 \ln x \, dx \]

### Step 3: Evaluate \( \int_0^1 \ln(1 - x) \, dx \)

Using integration by parts, let \( u = \ln(1 - x) \) and \( dv = dx \):
\[ du = -\frac{1}{1 - x} \, dx, \quad v = x \]
Thus,
\[ \int \ln(1 - x) \, dx = x \ln(1 - x) - \int \frac{x}{1 - x} \, dx \]
The second integral can be simplified further, but we can also use the known result:
\[ \int_0^1 \ln(1 - x) \, dx = -1 \]

### Step 4: Evaluate \( \int_0^1 \ln x \, dx \)

This integral is also a known result:
\[ \int_0^1 \ln x \, dx = -1 \]

### Step 5: Combine results for \( I_2 \)

Now substituting back into \( I_2 \):
\[ I_2 = (-1) - (-1) = 0 \]

### Conclusion

Both integrals \( I_1 \) and \( I_2 \) evaluate to 0:
\[ I_1 = 0, \quad I_2 = 0 \]
Thus, we conclude that:
\[ I_1 = I_2 \]

### Final Answer

The answer is \( I_1 = I_2 \).

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