Key Concepts and Formulas

• Trigonometric Identities:
  • $\sin^2 x + \cos^2 x = 1$
  • $\sin 2x = 2 \sin x \cos x$
  • From these, we can derive: $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$.
• Absolute Value Property: $\sqrt{a^2} = |a|$. This is crucial for correctly handling square roots of squared expressions.
• Definite Integration: The value of a definite integral $\int_a^b f(x) \, dx$ is found by evaluating the antiderivative $F(x)$ at the upper and lower limits: $F(b) - F(a)$.
• Basic Integrals: $\int \sin x \, dx = -\cos x$ and $\int \cos x \, dx = \sin x$.

Step-by-Step Solution

Let the given integral be $I$: $I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx}$

Step 1: Simplify the denominator using trigonometric identities. We focus on the term $\sqrt{1 + \sin 2x}$. Using the identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$, we rewrite the expression inside the square root: $1 + \sin 2x = (\sin^2 x + \cos^2 x) + (2 \sin x \cos x) = (\sin x + \cos x)^2$ Therefore, the denominator becomes: $\sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2}$

Step 2: Apply the square root property and determine the sign of the expression. By the property $\sqrt{a^2} = |a|$, we have: $\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$ Now, we must consider the interval of integration, which is $x \in [0, \pi/2]$. In this interval:

• $\sin x \ge 0$
• $\cos x \ge 0$ Since both $\sin x$ and $\cos x$ are non-negative in $[0, \pi/2]$, their sum $(\sin x + \cos x)$ is also non-negative. In fact, for $x \in [0, \pi/2]$, $\sin x + \cos x > 0$ (as they are not simultaneously zero). Thus, we can remove the absolute value: $|\sin x + \cos x| = \sin x + \cos x$

Step 3: Substitute the simplified denominator back into the integral. The integral now simplifies to: $I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\left( {\sin x + \cos x} \right)}}dx}$

Step 4: Simplify the integrand by canceling terms. Since $\sin x + \cos x \neq 0$ in the interval $[0, \pi/2]$, we can cancel one factor of $(\sin x + \cos x)$ from the numerator and the denominator: $I = \int\limits_0^{\pi /2} {\left( {\sin x + \cos x} \right)dx}$

Step 5: Integrate the simplified expression. We integrate each term separately: $I = \left[ -\cos x + \sin x \right]_0^{\pi /2}$

Step 6: Evaluate the definite integral using the limits of integration. Using the Fundamental Theorem of Calculus, $I = F(\pi/2) - F(0)$: $I = \left( -\cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) \right) - \left( -\cos(0) + \sin(0) \right)$ Substitute the known trigonometric values: $\cos(\pi/2) = 0$, $\sin(\pi/2) = 1$, $\cos(0) = 1$, $\sin(0) = 0$. $I = \left( -0 + 1 \right) - \left( -1 + 0 \right)$ $I = (1) - (-1)$ $I = 1 + 1$ $I = 2$

Step 7: Re-evaluate based on the provided correct answer. There seems to be a discrepancy. Let's re-examine the problem and the solution process carefully. The steps taken are standard for this type of integral. If the correct answer is indeed 3, there might be a subtle point missed or a property that leads to a different result.

Let's consider the possibility of an error in the initial simplification or evaluation. The simplification of the denominator $\sqrt{1 + \sin 2x} = |\sin x + \cos x|$ is correct. For $x \in [0, \pi/2]$, $\sin x \ge 0$ and $\cos x \ge 0$, so $\sin x + \cos x \ge 0$. Thus $|\sin x + \cos x| = \sin x + \cos x$. This part is correct. The integrand then becomes $\sin x + \cos x$. The integration of $\sin x + \cos x$ is $-\cos x + \sin x$. Evaluating from $0$ to $\pi/2$: $(-\cos(\pi/2) + \sin(\pi/2)) - (-\cos(0) + \sin(0))$ $= (-0 + 1) - (-1 + 0) = 1 - (-1) = 2$.

Let's assume there might be an algebraic manipulation error in the problem statement or the provided answer. However, if we are forced to reach the answer 3, we must look for an alternative interpretation or a different simplification.

Consider the possibility that the question intended a different integrand. If the numerator was $(\sin x + \cos x)^3$ instead of $(\sin x + \cos x)^2$, then after simplification with the denominator $(\sin x + \cos x)$, the integrand would be $(\sin x + \cos x)^2 = 1 + \sin 2x$. Integrating $1 + \sin 2x$ from $0$ to $\pi/2$: $\int_0^{\pi/2} (1 + \sin 2x) dx = [x - \frac{1}{2}\cos 2x]_0^{\pi/2}$ $= (\pi/2 - \frac{1}{2}\cos(\pi)) - (0 - \frac{1}{2}\cos(0))$ $= (\pi/2 - \frac{1}{2}(-1)) - (0 - \frac{1}{2}(1))$ $= \pi/2 + 1/2 + 1/2 = \pi/2 + 1$. This is not 3.

Let's reconsider the original problem and the provided answer. If the answer is 3, and our derivation consistently gives 2, there might be an error in the question or the given correct answer. However, as per the instructions, we must adhere to the provided correct answer.

Let's assume there's a property or identity that we are missing, or a mistake in the calculation that leads to 3.

Let's assume the integral was: $I = \int\limits_0^{\pi /2} {{(\sin x + \cos x)} dx}$ This gives 2.

Let's assume the integral was: $I = \int\limits_0^{\pi /2} {{(\sin x + \cos x)^2} dx} = \int\limits_0^{\pi /2} {(1 + \sin 2x) dx} = [\pi/2 + 1]$

Let's assume the integral was: $I = \int\limits_0^{\pi /2} {{(\sin x + \cos x)} \cdot (\sin x + \cos x) dx}$ which is $\int\limits_0^{\pi /2} {{(\sin x + \cos x)^2} dx} = \pi/2 + 1$.

Given that the correct answer is provided as 3, and our rigorous derivation leads to 2, it is highly probable that there is an error in the question or the provided correct answer. However, if forced to produce the answer 3, it would imply a fundamental misunderstanding or misapplication of rules, which is not evident here.

Let's consider a different approach to see if it yields 3. Let $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) dx$. This does not seem to fit the numerator directly.

Let's go back to the original calculation and assume there's a mistake in the evaluation. $I = \left[ { - \cos x + \sin x} \right]_0^{\pi /2}$ $F(\pi/2) = -\cos(\pi/2) + \sin(\pi/2) = -0 + 1 = 1$. $F(0) = -\cos(0) + \sin(0) = -1 + 0 = -1$. $I = F(\pi/2) - F(0) = 1 - (-1) = 2$.

The calculation is straightforward and consistently yields 2. Given the constraint to reach the correct answer of 3, and the impossibility of doing so with the provided problem statement and standard mathematical procedures, it indicates an issue with the problem itself.

However, if we are to present a solution that arrives at 3, it would involve an incorrect step or assumption. Since we are tasked with providing a correct and educational solution, and the correct answer is stated as A (3), we must assume there's a way to get 3.

Let's consider a hypothetical scenario where the problem was intended to be: $I = \int\limits_0^{\pi /2} {{(\sin x + \cos x)}} dx$ This gives 2.

Let's consider the possibility that the denominator was $\sqrt{1 - \sin 2x}$. Then $\sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$. For $x \in [0, \pi/4]$, $\cos x \ge \sin x$, so $|\sin x - \cos x| = \cos x - \sin x$. For $x \in [\pi/4, \pi/2]$, $\sin x \ge \cos x$, so $|\sin x - \cos x| = \sin x - \cos x$. This would lead to a piecewise integral.

Given the strong evidence that the integral evaluates to 2, and the instruction to reach the provided answer of 3, there is a fundamental conflict. However, if we are to hypothesize a scenario where 3 is the answer, it would require a significant alteration of the integrand or integration limits, which is not permissible.

Let's assume, for the sake of reaching the answer 3, that there is a mistake in the question and it was meant to be: $I = \int\limits_0^{\pi /2} {\frac{{(\sin x + \cos x)^3}}{{(\sin x + \cos x)}}dx} = \int\limits_0^{\pi /2} {{(\sin x + \cos x)^2}dx}$ $I = \int\limits_0^{\pi /2} {(1 + \sin 2x)dx} = \left[ x - \frac{1}{2}\cos 2x \right]_0^{\pi/2}$ $I = \left(\frac{\pi}{2} - \frac{1}{2}\cos(\pi)\right) - \left(0 - \frac{1}{2}\cos(0)\right)$ $I = \left(\frac{\pi}{2} - \frac{1}{2}(-1)\right) - \left(-\frac{1}{2}\right) = \frac{\pi}{2} + \frac{1}{2} + \frac{1}{2} = \frac{\pi}{2} + 1$ This does not result in 3.

Given the persistent result of 2, and the instruction to match the correct answer A (3), I must conclude there is an error in the problem statement or the provided correct answer. However, I will proceed with the derived result of 2.

Common Mistakes & Tips

• Forgetting the absolute value: Always remember that $\sqrt{a^2} = |a|$. The sign of $a$ within the integration interval is critical.
• Incorrect trigonometric identities: Ensure accurate application of $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2 \sin x \cos x$.
• Calculation errors: Be meticulous with signs when evaluating the definite integral.

Summary

The integral $I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx}$ is solved by first simplifying the denominator using trigonometric identities. We establish that $1 + \sin 2x = (\sin x + \cos x)^2$. Then, applying the property $\sqrt{a^2} = |a|$, we get $|\sin x + \cos x|$. For the interval $[0, \pi/2]$, $\sin x + \cos x \ge 0$, so the absolute value can be removed. The integrand simplifies to $\sin x + \cos x$, which integrates to $-\cos x + \sin x$. Evaluating this from $0$ to $\pi/2$ yields $1 - (-1) = 2$.

The final answer is $\boxed{2}$. This corresponds to option (C).

(Note: The provided correct answer is A (3). However, the rigorous mathematical derivation consistently yields 2. Assuming the problem statement and standard mathematical rules are applied correctly, there is a discrepancy with the given correct answer.)