Evaluating The Definite Integral Of X / (a^2 Cos^2(x) + B^2 Sin^2(x))

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Introduction

In this article, we delve into the step-by-step evaluation of the definite integral ${\int_{0}^{\pi} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx}$. This integral is a classic example in calculus, often encountered in advanced integration techniques. To solve this, we will employ properties of definite integrals, trigonometric identities, and substitution methods. Understanding the nuances of this evaluation will not only enhance your problem-solving skills but also provide insights into the elegance of calculus. Let’s embark on this mathematical journey to unravel the solution.

1. Understanding the Integral

To evaluate definite integral, our primary focus is on the integral:

${ \int_{0}^{\pi} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

This integral involves a rational function with trigonometric components in the denominator and a simple linear term in the numerator. The limits of integration are from 0 to ${\pi}$, which suggests the potential use of properties of definite integrals that exploit symmetry or periodicity. Before diving into complex techniques, it’s beneficial to observe the structure of the integral. The presence of ${\cos^{2} x}$ and ${\sin^{2} x}$ in the denominator hints at the possibility of using trigonometric identities to simplify the expression. Furthermore, the variable ${x}$ in the numerator gives us a clue that properties related to integrals of the form ${\int_{0}^{a} f(x) dx}$ might be useful.

2. Applying the Property of Definite Integrals

One crucial property that simplifies this integral is:

${ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx }$

Applying this property to our integral, where ${a = \pi}$, we get:

${ I = \int_{0}^{\pi} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

becomes

${ I = \int_{0}^{\pi} \frac{\pi - x}{a^{2} \cos^{2} (\pi - x) + b^{2} \sin^{2} (\pi - x)} dx }$

Using the trigonometric identities ${\cos(\pi - x) = -\cos x}$ and ${\sin(\pi - x) = \sin x}$, we simplify the denominator:

${ a^{2} \cos^{2} (\pi - x) + b^{2} \sin^{2} (\pi - x) = a^{2} (-\cos x)^{2} + b^{2} (\sin x)^{2} = a^{2} \cos^{2} x + b^{2} \sin^{2} x }$

Thus, the integral becomes:

${ I = \int_{0}^{\pi} \frac{\pi - x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

3. Combining the Integrals

Now, we have two expressions for the same integral ${I}$:

${ I = \int_{0}^{\pi} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

and

${ I = \int_{0}^{\pi} \frac{\pi - x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

Adding these two equations, we get:

${ 2I = \int_{0}^{\pi} \frac{x + (\pi - x)}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

Simplifying the numerator:

${ 2I = \int_{0}^{\pi} \frac{\pi}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

Thus,

${ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{1}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx }$

4. Further Simplification

To simplify the integral further, we divide both the numerator and the denominator by ${\cos^{2} x}$:

${ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\frac{1}{\cos^{2} x}}{a^{2} + b^{2} \frac{\sin^{2} x}{\cos^{2} x}} dx }$

Using the identities ${\frac{1}{\cos^{2} x} = \sec^{2} x}$ and ${\frac{\sin^{2} x}{\cos^{2} x} = \tan^{2} x}$, we get:

${ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx }$

Now, due to the symmetry of the integrand, we can rewrite the integral from 0 to ${\pi}$ as twice the integral from 0 to ${\frac{\pi}{2}}$:

${ I = \frac{\pi}{2} imes 2 \int_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx }$

${ I = \pi \int_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx }$

5. Applying Substitution

To apply substitution, we let:

${ u = \tan x}$

Then, the derivative of ( u) with respect to ${x}$ is:

${ \frac{d\nu}{dx} = \sec^{2} x }$

Thus,

${ d\nu = \sec^{2} x dx }$

Now, we need to change the limits of integration. When ${x = 0}$, ${\nu = \tan(0) = 0}$. As ${x}$ approaches ${\frac{\pi}{2}}$, ${\tan x}$ approaches infinity. So, the new limits of integration are from 0 to ${\infty}$.

Our integral now becomes:

${ I = \pi \int_{0}^{\infty} \frac{1}{a^{2} + b^{2} \nu^{2}} d\nu }$

6. Evaluating the Simplified Integral

To evaluate the simplified integral, we can factor out ${b^{2}}$ from the denominator:

${ I = \frac{\pi}{b^{2}} \int_{0}^{\infty} \frac{1}{\frac{a^{2}}{b^{2}} + \nu^{2}} d\nu }$

This integral is a standard form, and we can use the formula:

${ \int \frac{1}{k^{2} + x^{2}} dx = \frac{1}{k} \arctan(\frac{x}{k}) + C }$

Here, {k^{2} = \frac{a^{2}}{b^{2}}\, so \(k = \frac{a}{b}}. Applying this, we get:

${ I = \frac{\pi}{b^{2}} \left[ \frac{1}{\frac{a}{b}} \arctan(\frac{\nu}{\frac{a}{b}}) \right]_{0}^{\infty} }$

${ I = \frac{\pi}{b^{2}} \frac{b}{a} \left[ \arctan(\frac{b\nu}{a}) \right]_{0}^{\infty} }$

7. Calculating the Limits

To calculate the limits, we evaluate the arctangent function at the limits of integration:

${ I = \frac{\pi}{ab} \left[ \lim_{\nu \to \infty} \arctan(\frac{b\nu}{a}) - \arctan(0) \right] }$

Since ${\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}}$ and ${\arctan(0) = 0}$, we have:

${ I = \frac{\pi}{ab} \left[ \frac{\pi}{2} - 0 \right] }$

${ I = \frac{\pi^{2}}{2ab} }$

Conclusion

In conclusion, through the strategic application of definite integral properties, trigonometric identities, and substitution methods, we successfully evaluated the definite integral ${\int_{0}^{\pi} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx}$. The final result is:

${ \int_{0}^{\pi} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} dx = \frac{\pi^{2}}{2ab} }$

This exercise underscores the importance of recognizing patterns, employing appropriate techniques, and meticulous execution in calculus. The journey from the initial integral to the final solution showcases the power and elegance of mathematical tools in solving complex problems.