If `[int_0^1(dt)/(t^2+2tcosalpha+1)]x^2-[int_- 3^3(t^2sin2t)/(t^2+1)dt]x-2=0 (0 < alpha < pi)` then the value of x is

Options

A. (a)`pm(sqrt(alpha )/(2 sin alpha))`
B. (b)`pmsqrt((2 sin alpha )/(alpha))`
C. (c)`pmsqrt(( alpha)/( sin alpha))`
D. (d)`pm2sqrt((sin alpha)/(alpha))`

Answer

To solve the problem, we need to evaluate the integrals and then solve the resulting quadratic equation. Let's break down the solution step by step.

### Step 1: Evaluate the first integral \( I_1 \)

We have:
\[ I_1 = \int_0^1 \frac{dt}{t^2 + 2t \cos \alpha + 1} \]

This can be rewritten as:
\[ I_1 = \int_0^1 \frac{dt}{(t + \cos \alpha)^2 + \sin^2 \alpha} \]

This form is useful because it resembles the standard integral form for arctangent.

### Step 2: Use the substitution for the integral

Using the substitution \( u = t + \cos \alpha \), we have \( du = dt \) and the limits change from \( t = 0 \) to \( t = 1 \) which gives \( u = \cos \alpha \) to \( u = 1 + \cos \alpha \). Thus,
\[ I_1 = \int_{\cos \alpha}^{1 + \cos \alpha} \frac{du}{u^2 + \sin^2 \alpha} \]

### Step 3: Evaluate the integral

The integral \( \int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \). Here, \( a = \sin \alpha \):
\[ I_1 = \frac{1}{\sin \alpha} \left[ \tan^{-1} \left( \frac{1 + \cos \alpha}{\sin \alpha} \right) - \tan^{-1} \left( \frac{\cos \alpha}{\sin \alpha} \right) \right] \]

### Step 4: Simplify \( I_1 \)

Using the identity for the difference of arctangents:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \]
we can simplify this further.

### Step 5: Evaluate the second integral \( I_2 \)

Next, we evaluate:
\[ I_2 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt \]

Since \( t^2 \sin 2t \) is an odd function, the integral over a symmetric interval around zero is zero:
\[ I_2 = 0 \]

### Step 6: Set up the quadratic equation

Now substituting \( I_1 \) and \( I_2 \) back into the equation:
\[ I_1 x^2 - I_2 x - 2 = 0 \]
This simplifies to:
\[ I_1 x^2 - 2 = 0 \]

### Step 7: Solve for \( x \)

From the equation, we get:
\[ I_1 x^2 = 2 \implies x^2 = \frac{2}{I_1} \]
Taking the square root gives:
\[ x = \pm \sqrt{\frac{2}{I_1}} \]

### Step 8: Final expression for \( x \)

Thus, the value of \( x \) is:
\[ x = \pm \sqrt{\frac{2}{I_1}} \]