Key Concepts and Formulas

• Integral Equations: An equation where the unknown function appears under an integral sign.
• Properties of Definite Integrals:
  • A definite integral evaluates to a constant.
  • For an odd function $g(t)$ and a symmetric interval $[-a, a]$, $\int_{-a}^{a} g(t) dt = 0$.
  • For an even function $h(t)$ and a symmetric interval $[-a, a]$, $\int_{-a}^{a} h(t) dt = 2 \int_{0}^{a} h(t) dt$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.

Step-by-Step Solution

Step 1: Simplify the Integral Equation by Recognizing Constant Terms

We are given the integral equation: $f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt}$ The integral is with respect to $t$, so $\theta$ can be treated as a constant inside the integral. We can split the integral: $f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {\sin \theta f(t)dt} + \int\limits_{ - \pi /2}^{\pi /2} {t\cos \theta f(t)dt}$ $f(\theta ) = \sin \theta + \sin \theta \int\limits_{ - \pi /2}^{\pi /2} {f(t)dt} + \cos \theta \int\limits_{ - \pi /2}^{\pi /2} {t f(t)dt}$ Now, we can group the terms involving $\sin \theta$ and $\cos \theta$: $f(\theta ) = \sin \theta \left( 1 + \int\limits_{ - \pi /2}^{\pi /2} {f(t)dt} \right) + \cos \theta \left( \int\limits_{ - \pi /2}^{\pi /2} {t f(t)dt} \right)$ The definite integrals $\int\limits_{ - \pi /2}^{\pi /2} {f(t)dt}$ and $\int\limits_{ - \pi /2}^{\pi /2} {t f(t)dt}$ will evaluate to constant values.

Step 2: Define Constants and Assume a Form for $f(\theta)$

Let's define two constants: Let $A = 1 + \int\limits_{ - \pi /2}^{\pi /2} {f(t)dt}$ Let $B = \int\limits_{ - \pi /2}^{\pi /2} {t f(t)dt}$ Substituting these constants into the equation from Step 1, we get the form of $f(\theta)$: $f(\theta) = A \sin \theta + B \cos \theta$ This is a crucial step, as we have transformed the integral equation into a standard form for $f(\theta)$ with unknown constant coefficients.

Step 3: Substitute the Assumed Form of $f(\theta)$ Back into the Definitions of $A$ and $B$

Now, we substitute $f(t) = A \sin t + B \cos t$ into the definitions of $A$ and $B$ to form a system of equations for $A$ and $B$.

For Constant A: $A = 1 + \int\limits_{ - \pi /2}^{\pi /2} {(A \sin t + B \cos t)dt}$ $A = 1 + A \int\limits_{ - \pi /2}^{\pi /2} {\sin t dt} + B \int\limits_{ - \pi /2}^{\pi /2} {\cos t dt}$ We evaluate the integrals:

• $\int\limits_{ - \pi /2}^{\pi /2} {\sin t dt} = 0$ (since $\sin t$ is an odd function and the interval is symmetric).
• $\int\limits_{ - \pi /2}^{\pi /2} {\cos t dt} = [\sin t]_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2$. Substituting these values: $A = 1 + A(0) + B(2)$ $A = 1 + 2B \quad \text{... (i)}$

For Constant B: $B = \int\limits_{ - \pi /2}^{\pi /2} {t (A \sin t + B \cos t)dt}$ $B = A \int\limits_{ - \pi /2}^{\pi /2} {t \sin t dt} + B \int\limits_{ - \pi /2}^{\pi /2} {t \cos t dt}$ We evaluate these integrals:

• $\int\limits_{ - \pi /2}^{\pi /2} {t \cos t dt} = 0$ (since $t \cos t$ is an odd function and the interval is symmetric).
• $\int\limits_{ - \pi /2}^{\pi /2} {t \sin t dt}$: The function $t \sin t$ is an even function. So, $2 \int\limits_{ 0}^{\pi /2} {t \sin t dt}$. Using integration by parts with $u=t$ and $dv=\sin t \, dt$: $2 \left( [-t \cos t]_{0}^{\pi/2} - \int\limits_{ 0}^{\pi /2} {(-\cos t)dt} \right) = 2 \left( (0 - 0) + [\sin t]_{0}^{\pi/2} \right) = 2 (1 - 0) = 2$. Substituting these values: $B = A(2) + B(0)$ $B = 2A \quad \text{... (ii)}$

Step 4: Solve the System of Linear Equations for A and B

We have the system of equations: (i) $A = 1 + 2B$ (ii) $B = 2A$ Substitute (ii) into (i): $A = 1 + 2(2A)$ $A = 1 + 4A$ $-3A = 1 \implies A = -\frac{1}{3}$ Substitute the value of $A$ back into (ii): $B = 2 \left(-\frac{1}{3}\right) \implies B = -\frac{2}{3}$

Step 5: Determine the Explicit Form of $f(\theta)$

Using the values of $A$ and $B$ found in Step 4, the function $f(\theta)$ is: $f(\theta) = A \sin \theta + B \cos \theta = -\frac{1}{3} \sin \theta - \frac{2}{3} \cos \theta$ $f(\theta) = -\frac{1}{3} (\sin \theta + 2 \cos \theta)$

Step 6: Evaluate the Required Definite Integral

We need to find the value of $\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right|$. First, let's compute the integral $\int_0^{\pi /2} {f(\theta )d\theta }$: $\int_0^{\pi /2} {f(\theta )d\theta } = \int_0^{\pi /2} {-\frac{1}{3} (\sin \theta + 2 \cos \theta)d\theta }$ $= -\frac{1}{3} \int_0^{\pi /2} {(\sin \theta + 2 \cos \theta)d\theta }$ $= -\frac{1}{3} \left[ -\cos \theta + 2 \sin \theta \right]_0^{\pi/2}$ $= -\frac{1}{3} \left( (-\cos(\pi/2) + 2 \sin(\pi/2)) - (-\cos(0) + 2 \sin(0)) \right)$ $= -\frac{1}{3} \left( (0 + 2(1)) - (-1 + 0) \right)$ $= -\frac{1}{3} \left( 2 - (-1) \right)$ $= -\frac{1}{3} (3)$ $= -1$ Finally, we need to find the absolute value: $\left| \int_0^{\pi /2} {f(\theta )d\theta } \right| = |-1| = 1$

Common Mistakes & Tips

• Incorrectly evaluating integrals over symmetric intervals: Remember that integrals of odd functions over symmetric intervals are zero, and integrals of even functions can be simplified.
• Algebraic errors when solving for constants: Double-check your calculations when solving the system of linear equations for $A$ and $B$.
• Forgetting to take the absolute value at the end: The question specifically asks for the absolute value of the integral.

Summary

The problem involves solving an integral equation for the function $f(\theta)$. The key strategy is to recognize that the definite integrals within the equation are constants. By defining these constants and substituting the derived form of $f(\theta)$ back into their definitions, we create a system of linear equations. Solving this system yields the coefficients of $f(\theta)$. Finally, we evaluate the required definite integral of $f(\theta)$ and take its absolute value.

The final answer is $\boxed{1}$.