Key Concepts and Formulas

• Definite Integral Property (King's Property): For a definite integral $\int_a^b f(x) dx$, the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ holds. For the limits $0$ to $\pi$, this simplifies to $\int_0^\pi f(x) dx = \int_0^\pi f(\pi-x) dx$. This property is useful when substituting $a+b-x$ simplifies the integrand, especially when combined with the original integral.
• Trigonometric Identities: $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$. Consequently, $\cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x$.
• Substitution for Integration: If the integrand can be expressed in terms of a function and its derivative, a substitution can simplify the integration. For example, if the integral is of the form $\int g'(x) / g(x) dx$, the integral is $\ln|g(x)| + C$.

Step-by-Step Solution

Step 1: Define the integral and apply the King's Property. Let the given integral be $I$. $I = \int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \quad \text{..... (1)}$ We apply the property $\int_0^\pi f(x) dx = \int_0^\pi f(\pi-x) dx$. Replace $x$ with $(\pi-x)$ in the integrand: $I = \int_0^\pi \frac{((\pi-x)+3) \sin (\pi-x)}{1+3 \cos ^2 (\pi-x)} d x$ Using the trigonometric identities $\sin(\pi-x) = \sin x$ and $\cos^2(\pi-x) = \cos^2 x$, we get: $I = \int_0^\pi \frac{(\pi-x+3) \sin x}{1+3 \cos ^2 x} d x \quad \text{..... (2)}$

Step 2: Add the original integral and the transformed integral. Adding equation (1) and equation (2) allows us to combine the numerators over the common denominator. This step is crucial because the $x$ term in the numerator of equation (1) and the $-x$ term in the numerator of equation (2) will cancel out. $I + I = \int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x + \int_0^\pi \frac{(\pi-x+3) \sin x}{1+3 \cos ^2 x} d x$ $2I = \int_0^\pi \frac{(x+3) \sin x + (\pi-x+3) \sin x}{1+3 \cos ^2 x} d x$ Factor out $\sin x$ from the numerator: $2I = \int_0^\pi \frac{((x+3) + (\pi-x+3)) \sin x}{1+3 \cos ^2 x} d x$ Simplify the expression in the parentheses: $(x+3) + (\pi-x+3) = x+3+\pi-x+3 = \pi+6$. $2I = \int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos ^2 x} d x$

Step 3: Evaluate the simplified integral. We can take the constant $(\pi+6)$ out of the integral: $2I = (\pi+6) \int_0^\pi \frac{\sin x}{1+3 \cos ^2 x} d x$ To evaluate this integral, we use the substitution method. Let $u = \cos x$. Then, $du = -\sin x \, dx$, which means $\sin x \, dx = -du$. We also need to change the limits of integration: When $x = 0$, $u = \cos 0 = 1$. When $x = \pi$, $u = \cos \pi = -1$. Substitute these into the integral: $2I = (\pi+6) \int_1^{-1} \frac{-du}{1+3u^2}$ We can reverse the limits of integration by changing the sign of the integral: $2I = (\pi+6) \int_{-1}^{1} \frac{du}{1+3u^2}$ To integrate $\frac{1}{1+3u^2}$, we can rewrite it as $\frac{1}{3} \frac{1}{(1/\sqrt{3})^2 + u^2}$. Recall the standard integral formula $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$. Here, $a = 1/\sqrt{3}$. $\int \frac{du}{1+3u^2} = \frac{1}{3} \int \frac{du}{(1/\sqrt{3})^2 + u^2} = \frac{1}{3} \left( \frac{1}{1/\sqrt{3}} \arctan\left(\frac{u}{1/\sqrt{3}}\right) \right) = \frac{1}{3} \left( \sqrt{3} \arctan(\sqrt{3}u) \right) = \frac{1}{\sqrt{3}} \arctan(\sqrt{3}u)$ Now, evaluate the definite integral: $2I = (\pi+6) \left[ \frac{1}{\sqrt{3}} \arctan(\sqrt{3}u) \right]_{-1}^{1}$ $2I = (\pi+6) \frac{1}{\sqrt{3}} \left( \arctan(\sqrt{3} \cdot 1) - \arctan(\sqrt{3} \cdot (-1)) \right)$ $2I = \frac{\pi+6}{\sqrt{3}} \left( \arctan(\sqrt{3}) - \arctan(-\sqrt{3}) \right)$ We know that $\arctan(\sqrt{3}) = \frac{\pi}{3}$ and $\arctan(-\sqrt{3}) = -\frac{\pi}{3}$. $2I = \frac{\pi+6}{\sqrt{3}} \left( \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) \right)$ $2I = \frac{\pi+6}{\sqrt{3}} \left( \frac{\pi}{3} + \frac{\pi}{3} \right)$ $2I = \frac{\pi+6}{\sqrt{3}} \left( \frac{2\pi}{3} \right)$

Step 4: Solve for I. $2I = \frac{2\pi(\pi+6)}{3\sqrt{3}}$ Divide both sides by 2: $I = \frac{\pi(\pi+6)}{3\sqrt{3}}$ This can be written as: $I = \frac{\pi}{3\sqrt{3}}(\pi+6)$

Common Mistakes & Tips

• Incorrect application of the King's Property: Ensure that the substitution $a+b-x$ is correctly applied to all parts of the integrand.
• Errors in trigonometric simplification: Double-check identities like $\sin(\pi-x)$ and $\cos(\pi-x)$, especially when squared.
• Mistakes in substitution: When changing the limits of integration during substitution, ensure they correspond to the new variable. Also, be careful with the sign when $du$ is related to $-f'(x)dx$.
• Incorrect arctan values: Remember the principal values of arctan for common angles.

Summary

The integral was solved using the King's Property, which transforms the integral by replacing $x$ with $\pi-x$. Adding the original and transformed integrals simplified the integrand by cancelling out the $x$ term in the numerator. The resulting integral was evaluated using a trigonometric substitution ($u = \cos x$) and standard integration formulas for arctan. The final result was obtained by solving for $I$.

The final answer is $\boxed{\frac{\pi}{3 \sqrt{3}}(\pi+6)}$.