If `I_(1)= int_(1)^(sin theta) (x)/(1+x^(2)) dx and I_(2) int_(1)^("cosec" theta) (1)/(x(x^(2+1)))dx` then the value of `|{:(I_(1),I_(1)^(2),I_(2)),(e^(I_(1)+I_(2)),I_(2)^(2),-1),(1,I_(1)^(2)+I_(2)^(2),-1):}|` , is

Options

A. `sin theta`
B. `"cosec" theta`
C. 0
D. 1

Explanation

To solve the given problem, we need to evaluate the determinant involving the integrals \( I_1 \) and \( I_2 \). Let's break down the steps systematically.

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

The integral \( I_1 \) is defined as:
\[ I_1 = \int_{1}^{\sin \theta} \frac{x}{1+x^2} \, dx \]

To solve this integral, we can use a substitution. The derivative of \( 1 + x^2 \) is \( 2x \), which suggests that we can use the substitution \( u = 1 + x^2 \). Thus, \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \).

However, a simpler approach is to recognize that:
\[ \frac{d}{dx}(\ln(1+x^2)) = \frac{2x}{1+x^2} \]
This means:
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C \]

Now, we can evaluate \( I_1 \):
\[ I_1 = \left[ \frac{1}{2} \ln(1+x^2) \right]_{1}^{\sin \theta} \]
Calculating the limits:
\[ I_1 = \frac{1}{2} \ln(1+\sin^2 \theta) - \frac{1}{2} \ln(2) = \frac{1}{2} \left( \ln(1+\sin^2 \theta) - \ln(2) \right) = \frac{1}{2} \ln\left(\frac{1+\sin^2 \theta}{2}\right) \]

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

The integral \( I_2 \) is defined as:
\[ I_2 = \int_{1}^{\csc \theta} \frac{1}{x(x^2+1)} \, dx \]

We can simplify this integral using partial fractions:
\[ \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \]

Multiplying through by the denominator \( x(x^2+1) \) and solving for \( A, B, C \):
\[ 1 = A(x^2 + 1) + (Bx + C)x \]
Setting \( x = 0 \) gives \( A = 1 \). By substituting \( x = 1 \) and \( x = -1 \), we can find \( B \) and \( C \).

After finding the coefficients, we can integrate:
\[ I_2 = \int \left( \frac{1}{x} - \frac{x}{x^2 + 1} \right) \, dx \]
This results in:
\[ I_2 = \ln |x| - \frac{1}{2} \ln(x^2 + 1) + C \]

Evaluating from 1 to \( \csc \theta \):
\[ I_2 = \left[ \ln |x| - \frac{1}{2} \ln(x^2 + 1) \right]_{1}^{\csc \theta} \]
Calculating the limits gives:
\[ I_2 = \left( \ln(\csc \theta) - \frac{1}{2} \ln(\csc^2 \theta + 1) \right) - \left( \ln(1) - \frac{1}{2} \ln(2) \right) \]
This simplifies to:
\[ I_2 = \ln(\csc \theta) - \frac{1}{2} \ln(\csc^2 \theta + 1) + \frac{1}{2} \ln(2) \]

### Step 3: Relationship Between \( I_1 \) and \( I_2 \)

From the video transcript, we find that:
\[ I_2 = -I_1 \]
Thus, we can substitute \( I_2 \) in terms of \( I_1 \) in our determinant.

### Step 4: Setting Up the Determinant

The determinant we need to evaluate is:
\[ D = \begin{vmatrix} I_1 & I_1^2 & I_2 \\ e^{I_1 + I_2} & I_2^2 & -1 \\ 1 & I_1^2 + I_2^2 & -1 \end{vmatrix} \]

Substituting \( I_2 = -I_1 \):
\[ D = \begin{vmatrix} I_1 & I_1^2 & -I_1 \\ e^0 & (-I_1)^2 & -1 \\ 1 & I_1^2 + (-I_1)^2 & -1 \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} I_1 & I_1^2 & -I_1 \\ 1 & I_1^2 & -1 \\ 1 & 2I_1^2 & -1 \end{vmatrix} \]

### Step 5: Evaluating the Determinant

Notice that the second and third rows are linearly dependent (the third row is a scalar multiple of the second row). Therefore, the value of the determinant is:
\[ D = 0 \]

### Final Answer

Thus, the value of the determinant is:
\[ \boxed{0} \]