Key Concepts and Formulas

• King's Property of Definite Integrals: For an integral with limits $[a, b]$, the following property holds: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ This property is particularly useful for integrals with symmetric limits or when the integrand simplifies upon the substitution $x \to a+b-x$.
• Algebraic Manipulation of Integrals: If we have two forms of the same integral, say $I = \int_a^b f(x) dx$ and $I = \int_a^b g(x) dx$, we can add or subtract them: $2I = \int_a^b (f(x) + g(x)) dx$ or $0 = \int_a^b (f(x) - g(x)) dx$.
• Trigonometric Identity: The double angle formula for cosine in terms of tangent: $\cos(2x) = \frac{1 - \tan^2 x}{1 + \tan^2 x}$
• Standard Integral Formula: $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ More generally, for $a^2+u^2$: $\int \frac{1}{a^2 + u^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$

Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx} \quad \cdots (1)$

Step 1: Apply 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 $b = \frac{\pi}{4}$. So, $a+b = -\frac{\pi}{4} + \frac{\pi}{4} = 0$. Thus, we replace $x$ with $a+b-x = 0-x = -x$ in the integrand. The denominator involves $\cos(2x)$, and since $\cos(2(-x)) = \cos(-2x) = \cos(2x)$, the denominator remains unchanged. The numerator changes from $x + \frac{\pi}{4}$ to $(-x) + \frac{\pi}{4}$. $I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{(-x) + {\pi \over 4}} \over {2 - \cos (2(-x))}}dx}$ $I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{-x + {\pi \over 4}} \over {2 - \cos 2x}}dx} \quad \cdots (2)$ Reasoning: This step is crucial for simplifying the numerator. By creating a second form of the integral, we can combine it with the original integral to eliminate the $x$ term in the numerator.

Step 2: Combine the two forms of the integral Add equation (1) and equation (2): $I + I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx} + \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{-x + {\pi \over 4}} \over {2 - \cos 2x}}dx}$ $2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{(x + {\pi \over 4}) + (-x + {\pi \over 4})} \over {2 - \cos 2x}}dx}$ $2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{2 {\pi \over 4}} \over {2 - \cos 2x}}dx}$ $2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{\pi \over 2}} \over {2 - \cos 2x}}dx$ $I = {1 \over 2} \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{\pi \over 2}} \over {2 - \cos 2x}}dx$ $I = {\pi \over 4} \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{1} \over {2 - \cos 2x}}dx$ Reasoning: Adding the two integral forms allows us to combine the numerators, which simplifies the expression significantly by canceling out the $x$ terms.

Step 3: Simplify the integrand using trigonometric identities We use the identity $\cos(2x) = \frac{1 - \tan^2 x}{1 + \tan^2 x}$. Substitute this into the integrand: $\frac{1}{2 - \cos 2x} = \frac{1}{2 - \frac{1 - \tan^2 x}{1 + \tan^2 x}}$ $= \frac{1 + \tan^2 x}{2(1 + \tan^2 x) - (1 - \tan^2 x)}$ $= \frac{1 + \tan^2 x}{2 + 2\tan^2 x - 1 + \tan^2 x}$ $= \frac{1 + \tan^2 x}{1 + 3\tan^2 x}$ Recall that $1 + \tan^2 x = \sec^2 x$. So the integrand becomes: $\frac{\sec^2 x}{1 + 3\tan^2 x}$ Now, the integral is: $I = {\pi \over 4} \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\sec^2 x \over {1 + 3\tan^2 x}}dx$ Reasoning: Expressing $\cos(2x)$ in terms of $\tan^2 x$ is a common strategy when the integral involves $\sec^2 x$ in the numerator, as it allows for a substitution involving $\tan x$.

Step 4: Perform a substitution Let $u = \tan x$. Then, $du = \sec^2 x \, dx$. We also need to change the limits of integration. When $x = -\frac{\pi}{4}$, $u = \tan(-\frac{\pi}{4}) = -1$. When $x = \frac{\pi}{4}$, $u = \tan(\frac{\pi}{4}) = 1$. The integral becomes: $I = {\pi \over 4} \int\limits_{-1}^{1} {1 \over {1 + 3u^2}}du$ Reasoning: This substitution simplifies the integrand to a standard form that can be easily integrated.

Step 5: Evaluate the integral with respect to $u$ We can rewrite the denominator as $1^2 + (\sqrt{3}u)^2$. $I = {\pi \over 4} \int\limits_{-1}^{1} {1 \over {1^2 + (\sqrt{3}u)^2}}du$ To match the standard form $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$, we need to adjust for the coefficient of $u$. Let $v = \sqrt{3}u$. Then $dv = \sqrt{3} du$, so $du = \frac{1}{\sqrt{3}} dv$. The limits of integration also change: When $u = -1$, $v = \sqrt{3}(-1) = -\sqrt{3}$. When $u = 1$, $v = \sqrt{3}(1) = \sqrt{3}$. $I = {\pi \over 4} \int\limits_{-\sqrt{3}}^{\sqrt{3}} {1 \over {1^2 + v^2}} \left(\frac{1}{\sqrt{3}} dv\right)$ $I = {\pi \over 4\sqrt{3}} \int\limits_{-\sqrt{3}}^{\sqrt{3}} {1 \over {1^2 + v^2}} dv$ Now, apply the standard integral formula $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$ with $a=1$ and variable $v$. $I = {\pi \over 4\sqrt{3}} \left[ {1 \over 1} \tan^{-1}\left(\frac{v}{1}\right) \right]_{-\sqrt{3}}^{\sqrt{3}}$ $I = {\pi \over 4\sqrt{3}} \left[ \tan^{-1}(v) \right]_{-\sqrt{3}}^{\sqrt{3}}$ Reasoning: This step involves applying a standard integration formula after a suitable substitution and adjustment for the coefficient.

Step 6: Evaluate the definite integral $I = {\pi \over 4\sqrt{3}} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(-\sqrt{3}) \right)$ We know that $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ and $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$ (since $\tan^{-1}$ is an odd function). $I = {\pi \over 4\sqrt{3}} \left( \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) \right)$ $I = {\pi \over 4\sqrt{3}} \left( \frac{\pi}{3} + \frac{\pi}{3} \right)$ $I = {\pi \over 4\sqrt{3}} \left( \frac{2\pi}{3} \right)$ $I = {\pi \over {2\sqrt{3}}} \left( \frac{\pi}{3} \right)$ $I = {{\pi^2} \over {6\sqrt 3 }}$ Reasoning: Evaluating the $\tan^{-1}$ function at the limits and simplifying the expression yields the final value of the integral.

Common Mistakes & Tips

• Incorrect application of King's Property: Ensure that $a+b-x$ is correctly substituted into the entire integrand, especially in the numerator.
• Trigonometric identity errors: Double-check the conversion of $\cos(2x)$ to $\tan^2 x$ and the subsequent algebraic simplification.
• Substitution errors: Be careful with changing the limits of integration and adjusting for any coefficients introduced during substitution (e.g., the $\sqrt{3}$ in $v = \sqrt{3}u$).

Summary

The integral was solved by first applying King's Property to generate a second form of the integral. Combining the original and transformed integrals simplified the integrand significantly. Subsequently, the trigonometric identity for $\cos(2x)$ in terms of $\tan^2 x$ was used, followed by a substitution $u = \tan x$. This transformed the integral into a standard form involving $\tan^{-1}$, which was then evaluated using the fundamental theorem of calculus. The final result obtained is $\frac{\pi^2}{6\sqrt 3}$.

The final answer is $\boxed{{\frac{\pi^2}{6\sqrt 3}}}$.