Let `f(x)=|(1,a,a^2),(sin(n-1)x,sinnx,sin(n+1)x),(cos(n-1)x,cosnx,cos(n+1)x)|` then `int_(0)^(pi//2)` f (x) dx is equal to

Options

A. `a-(1+a^2)`
B. `1+a+a^2`
C. `-a+(1+a^2)`
D. `-(1+a+a^2)`

Answer

To solve the given problem, we need to evaluate the integral \( \int_{0}^{\frac{\pi}{2}} f(x) \, dx \) where

\[ f(x) = \left| \begin{array}{ccc} 1 & a & a^2 \\ \sin((n-1)x) & \sin(nx) & \sin((n+1)x) \\ \cos((n-1)x) & \cos(nx) & \cos((n+1)x) \end{array} \right| \]

### Step 1: Calculate the Determinant

We will calculate the determinant \( f(x) \) using the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ f(x) = 1 \cdot \left( \sin(nx) \cdot \cos((n+1)x) - \sin((n+1)x) \cdot \cos(nx) \right) - a \cdot \left( \sin((n-1)x) \cdot \cos((n+1)x) - \sin((n+1)x) \cdot \cos((n-1)x) \right) + a^2 \cdot \left( \sin((n-1)x) \cdot \cos(nx) - \sin(nx) \cdot \cos((n-1)x) \right) \]

### Step 2: Simplify Each Term

Using the identity \( \sin A \cos B - \cos A \sin B = \sin(A - B) \), we can simplify each term:

1. The first term simplifies to:
\[ \sin(nx) \cos((n+1)x) - \sin((n+1)x) \cos(nx) = \sin(nx - (n+1)x) = \sin(-x) = -\sin(x) \]

2. The second term simplifies to:
\[ -a \left( \sin((n-1)x) \cos((n+1)x) - \sin((n+1)x) \cos((n-1)x) \right) = -a \sin((n-1)x - (n+1)x) = -a \sin(-2x) = a \sin(2x) \]

3. The third term simplifies to:
\[ a^2 \left( \sin((n-1)x) \cos(nx) - \sin(nx) \cos((n-1)x) \right) = a^2 \sin((n-1)x - nx) = a^2 \sin(-x) = -a^2 \sin(x) \]

### Step 3: Combine the Terms

Combining all the simplified terms, we have:

\[ f(x) = -\sin(x) + a \sin(2x) - a^2 \sin(x) = (a \sin(2x) - (1 + a^2) \sin(x)) \]

### Step 4: Set Up the Integral

Now we need to evaluate the integral:

\[ \int_{0}^{\frac{\pi}{2}} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} \left( a \sin(2x) - (1 + a^2) \sin(x) \right) \, dx \]

### Step 5: Evaluate the Integral

1. The integral of \( \sin(2x) \) is:
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) \]
Therefore,
\[ \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{2} (-1 - 1) = 1 \]

2. The integral of \( \sin(x) \) is:
\[ \int \sin(x) \, dx = -\cos(x) \]
Therefore,
\[ \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = \left[-\cos(x)\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

### Step 6: Combine Results

Now substituting back into the integral:

\[ \int_{0}^{\frac{\pi}{2}} f(x) \, dx = a \cdot 1 - (1 + a^2) \cdot 1 = a - (1 + a^2) = a - 1 - a^2 \]

### Final Answer

Thus, the final result is:

\[ \int_{0}^{\frac{\pi}{2}} f(x) \, dx = a - 1 - a^2 \]