Key Concepts and Formulas

• Definite Integral Property (King's Property): For a definite integral $\int_a^b f(x) dx$, if $a+b = k$ (a constant), then $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. This property is extremely useful for simplifying integrals, especially when the integrand contains $x$ in a way that can be transformed.
• Trigonometric Identities:
  • $\sin(\pi - \theta) = \sin \theta$
  • $\sin(\frac{\pi}{2} + \theta) = \cos \theta$
  • $1 + \cos(2\theta) = 2\cos^2 \theta$
  • $\sin(2\theta) = 2\sin \theta \cos \theta$
• Integration of $\sec x$ and $\csc x$:
  • $\int \sec x dx = \ln|\sec x + \tan x| + C$
  • $\int \csc x dx = \ln|\csc x - \cot x| + C$ or $\ln|\tan(\frac{x}{2})| + C$

Step-by-Step Solution

Let the given integral be $I$. $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx}$

Step 1: Apply the King's Property. We use the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a = \frac{\pi}{4}$ and $b = \frac{3\pi}{4}$. So, $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \pi$. Let $f(x) = \frac{x}{1 + \sin x}$. Then $f(a+b-x) = f(\pi-x) = \frac{\pi-x}{1 + \sin(\pi-x)}$. Since $\sin(\pi-x) = \sin x$, we have: $f(\pi-x) = \frac{\pi-x}{1 + \sin x}$ Applying the property, we get: $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{\pi-x}{1 + \sin x}}dx}$

Step 2: Add the original integral and the transformed integral. We have two expressions for $I$: $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx} \quad (*)$ $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{\pi-x}{1 + \sin x}}dx} \quad (**)$ Adding $(*)$ and $(**)$: $2I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{x}{1 + \sin x}}dx} + \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{\pi-x}{1 + \sin x}}dx}$ Since the limits of integration are the same, we can combine the integrands: $2I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{x + (\pi-x)}{1 + \sin x}}dx}$ $2I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{\pi}{1 + \sin x}}dx}$ $2I = \pi \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{\frac{1}{1 + \sin x}}dx}$

Step 3: Simplify the integrand $\frac{1}{1 + \sin x}$. To integrate $\frac{1}{1 + \sin x}$, we can multiply the numerator and denominator by $1 - \sin x$: $\frac{1}{1 + \sin x} = \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} = \frac{1 - \sin x}{1 - \sin^2 x}$ Using the identity $1 - \sin^2 x = \cos^2 x$: $\frac{1 - \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}$ $\frac{1}{\cos^2 x} - \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \sec^2 x - \tan x \sec x$

Step 4: Integrate the simplified expression. Now we need to integrate $\sec^2 x - \sec x \tan x$: $\int (\sec^2 x - \sec x \tan x) dx = \int \sec^2 x dx - \int \sec x \tan x dx$ The integral of $\sec^2 x$ is $\tan x$, and the integral of $\sec x \tan x$ is $\sec x$. So, the antiderivative is $\tan x - \sec x$.

Step 5: Evaluate the definite integral. Now we evaluate the definite integral of $\sec^2 x - \sec x \tan x$ from $\frac{\pi}{4}$ to $\frac{3\pi}{4}$: $2I = \pi \left[ \tan x - \sec x \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$ $2I = \pi \left[ \left( \tan \frac{3\pi}{4} - \sec \frac{3\pi}{4} \right) - \left( \tan \frac{\pi}{4} - \sec \frac{\pi}{4} \right) \right]$

Let's evaluate the trigonometric terms:

• $\tan \frac{3\pi}{4} = -1$
• $\sec \frac{3\pi}{4} = \frac{1}{\cos \frac{3\pi}{4}} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2}$
• $\tan \frac{\pi}{4} = 1$
• $\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$

Substitute these values back into the expression for $2I$: $2I = \pi \left[ (-1 - (-\sqrt{2})) - (1 - \sqrt{2}) \right]$ $2I = \pi \left[ (-1 + \sqrt{2}) - (1 - \sqrt{2}) \right]$ $2I = \pi \left[ -1 + \sqrt{2} - 1 + \sqrt{2} \right]$ $2I = \pi \left[ 2\sqrt{2} - 2 \right]$ $2I = 2\pi (\sqrt{2} - 1)$

Step 6: Solve for I. Divide both sides by 2: $I = \pi (\sqrt{2} - 1)$

Alternative approach for Step 3: We can also rewrite $1+\sin x$ using half-angle formulas. Let $x = \frac{\pi}{2} + \theta$. Then $\sin x = \sin(\frac{\pi}{2} + \theta) = \cos \theta$. The limits change: When $x = \frac{\pi}{4}$, $\theta = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4}$. When $x = \frac{3\pi}{4}$, $\theta = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$. So, $dx = d\theta$. The integral becomes: $I = \int_{-\pi/4}^{\pi/4} \frac{\frac{\pi}{2}+\theta}{1+\cos\theta} d\theta$ $I = \int_{-\pi/4}^{\pi/4} \frac{\frac{\pi}{2}+\theta}{2\cos^2(\theta/2)} d\theta$ $I = \frac{1}{2} \int_{-\pi/4}^{\pi/4} (\frac{\pi}{2}+\theta) \sec^2(\theta/2) d\theta$ This approach seems more complicated due to the $\theta$ term in the numerator. The first method of multiplying by the conjugate is more straightforward here.

Let's re-verify the integration of $\frac{1}{1+\sin x}$ using a different identity if needed. We can also write $1+\sin x = 1 + \cos(\frac{\pi}{2}-x) = 2\cos^2(\frac{\pi}{4}-\frac{x}{2})$. So, $\frac{1}{1+\sin x} = \frac{1}{2\cos^2(\frac{\pi}{4}-\frac{x}{2})} = \frac{1}{2}\sec^2(\frac{\pi}{4}-\frac{x}{2})$. Let $u = \frac{\pi}{4}-\frac{x}{2}$. Then $du = -\frac{1}{2} dx$, so $dx = -2 du$. When $x = \frac{\pi}{4}$, $u = \frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$. When $x = \frac{3\pi}{4}$, $u = \frac{\pi}{4} - \frac{3\pi}{8} = -\frac{\pi}{8}$. The integral becomes: $\int \frac{1}{1+\sin x} dx = \int \frac{1}{2}\sec^2(u) (-2 du) = -\int \sec^2(u) du = -\tan(u) + C$ $= -\tan(\frac{\pi}{4}-\frac{x}{2}) + C$ Now evaluate from $\frac{\pi}{4}$ to $\frac{3\pi}{4}$: $2I = \pi \left[ -\tan(\frac{\pi}{4}-\frac{x}{2}) \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$ $2I = \pi \left[ -\tan(\frac{\pi}{4}-\frac{3\pi}{8}) - (-\tan(\frac{\pi}{4}-\frac{\pi}{8})) \right]$ $2I = \pi \left[ -\tan(-\frac{\pi}{8}) + \tan(\frac{\pi}{8}) \right]$ Since $\tan(-\theta) = -\tan(\theta)$: $2I = \pi \left[ \tan(\frac{\pi}{8}) + \tan(\frac{\pi}{8}) \right] = 2\pi \tan(\frac{\pi}{8})$ We need to find $\tan(\frac{\pi}{8})$. We know $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$. Let $\theta = \frac{\pi}{8}$, so $2\theta = \frac{\pi}{4}$. $\tan(\frac{\pi}{4}) = 1$. $1 = \frac{2\tan(\frac{\pi}{8})}{1-\tan^2(\frac{\pi}{8})}$. Let $t = \tan(\frac{\pi}{8})$. $1 = \frac{2t}{1-t^2}$ $1-t^2 = 2t$ $t^2 + 2t - 1 = 0$. Using the quadratic formula, $t = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4+4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$. Since $\frac{\pi}{8}$ is in the first quadrant, $\tan(\frac{\pi}{8})$ must be positive. So, $\tan(\frac{\pi}{8}) = \sqrt{2} - 1$. Substituting this back into $2I = 2\pi \tan(\frac{\pi}{8})$: $2I = 2\pi (\sqrt{2} - 1)$ $I = \pi (\sqrt{2} - 1)$ This confirms the result obtained by the first method.

Common Mistakes & Tips

• Incorrect application of King's Property: Ensure that the sum $a+b$ is indeed a constant and that the transformation $f(a+b-x)$ simplifies the integral.
• Errors in trigonometric values: Be careful when evaluating trigonometric functions at specific angles, especially in the second quadrant (like $\frac{3\pi}{4}$). Double-check signs and values.
• Algebraic mistakes during integration or evaluation: Simplification steps, especially with fractions and square roots, can lead to errors. Always review your calculations.
• Not recognizing the simplification of $\frac{1}{1+\sin x}$: This is a common integrand in definite integrals. Knowing to multiply by the conjugate or use half-angle identities is crucial.

Summary

The problem involves evaluating a definite integral with limits $\frac{\pi}{4}$ and $\frac{3\pi}{4}$. The key to solving this integral is to utilize the King's property of definite integrals, which states $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. By applying this property with $a+b = \pi$, we transform the integral into a form where adding the original and transformed integrals significantly simplifies the numerator to a constant $\pi$. The resulting integral of $\frac{1}{1+\sin x}$ can be solved by multiplying the numerator and denominator by the conjugate $(1-\sin x)$ to obtain terms involving $\sec^2 x$ and $\sec x \tan x$, which are standard integrals. Evaluating these at the given limits leads to the final answer.

The final answer is \boxed{\pi \left( {\sqrt 2 - 1} \right)}.