Evaluating The Definite Integral Of ∫[0 To 15] Dx/√(324 + X^2)

by THE IDEN 63 views

Introduction

In this comprehensive article, we will delve into the evaluation of the definite integral $\int_0^{15} \frac{dx}{\sqrt{324 + x^2}}$ using trigonometric substitution. Integrals of this form frequently appear in calculus and require a solid understanding of trigonometric identities and substitution techniques. Our step-by-step approach will ensure clarity and precision, making the process accessible even to those who are relatively new to integral calculus. This detailed guide aims to provide not just the solution, but also a thorough explanation of the methods involved. By the end of this article, you will be well-equipped to tackle similar integrals and have a deeper appreciation for the elegance of calculus.

Understanding the Integral

The integral we aim to solve is a definite integral, which means we are looking for the area under the curve of the function $f(x) = \frac{1}{\sqrt{324 + x^2}}$ between the limits of integration 0 and 15. The integrand contains a square root term in the denominator, specifically of the form $\sqrt{a^2 + x^2}\$ where $a^2 = 324$. This form suggests the use of a trigonometric substitution to simplify the integral. Trigonometric substitution is a powerful technique used to simplify integrals involving square roots of quadratic expressions. By substituting the variable $x$ with a trigonometric function, we can often eliminate the square root and transform the integral into a more manageable form. The key to successful trigonometric substitution lies in choosing the appropriate substitution based on the form of the integrand. In this case, the presence of $\sqrt{a^2 + x^2}\$ guides us to use the tangent substitution.

Trigonometric Substitution

The technique of trigonometric substitution is particularly useful when dealing with integrals containing expressions of the form $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$. Each of these forms suggests a different trigonometric substitution that can simplify the integrand. For our integral, which contains $\sqrt{324 + x^2}$, we will use the substitution $x = a \tan(\theta)$, where $a^2 = 324$. This substitution is motivated by the trigonometric identity $1 + \tan^2(\theta) = \sec^2(\theta)$, which allows us to eliminate the square root. The choice of $\tan(\theta)$ is crucial because it aligns perfectly with the structure of our integral, making the subsequent steps much smoother. Understanding why this substitution works is essential for mastering this technique. The goal is to transform the integrand into a form that can be easily integrated using standard trigonometric integrals.

Step-by-Step Solution

1. Trigonometric Substitution

To evaluate the integral $\int_0^{15} \frac{dx}{\sqrt{324 + x^2}}$ we start by making a trigonometric substitution. Let $x = 18 \tan(\theta)$, where $18 = \sqrt{324}$. This implies that $dx = 18 \sec^2(\theta) d\theta$. This substitution is crucial as it will help us eliminate the square root in the denominator. The derivative of $x$ with respect to $\theta$ gives us the necessary transformation for $dx$. This step is the cornerstone of the entire process, and a clear understanding of this substitution is vital for proceeding further. By replacing $x$ with a trigonometric function, we are effectively changing the variable of integration, which will simplify the integral significantly.

2. Substitute into the Integral

Next, we substitute $x = 18 \tan(\theta)$ and $dx = 18 \sec^2(\theta) d\theta$ into the integral:

∫dx324+x2=∫18sec⁡2(θ)dθ324+(18tan⁡(θ))2\int \frac{dx}{\sqrt{324 + x^2}} = \int \frac{18 \sec^2(\theta) d\theta}{\sqrt{324 + (18 \tan(\theta))^2}}

This substitution transforms the integral into a trigonometric form. The goal here is to rewrite the integral in terms of $\theta$, which should be easier to evaluate. This step involves careful replacement of both $x$ and $dx$ with their equivalent trigonometric expressions. The substitution is not just about replacing variables; it's about transforming the entire integral into a new domain where it can be solved more easily. Accuracy in this step is paramount, as any error here will propagate through the rest of the solution.

3. Simplify the Expression

Now, we simplify the expression inside the integral:

∫18sec⁡2(θ)dθ324+324tan⁡2(θ)=∫18sec⁡2(θ)dθ324(1+tan⁡2(θ))\int \frac{18 \sec^2(\theta) d\theta}{\sqrt{324 + 324 \tan^2(\theta)}} = \int \frac{18 \sec^2(\theta) d\theta}{\sqrt{324(1 + \tan^2(\theta))}}

We can factor out 324 from the square root. This simplification is a critical step, as it allows us to use trigonometric identities to further reduce the complexity of the integral. The ability to simplify expressions is a key skill in integral calculus, and this step exemplifies its importance. By factoring out constants and applying trigonometric identities, we move closer to a form that can be directly integrated.

4. Apply Trigonometric Identity

Using the trigonometric identity $1 + \tan^2(\theta) = \sec^2(\theta)$, we further simplify the expression:

∫18sec⁡2(θ)dθ324sec⁡2(θ)=∫18sec⁡2(θ)dθ18sec⁡(θ)\int \frac{18 \sec^2(\theta) d\theta}{\sqrt{324 \sec^2(\theta)}} = \int \frac{18 \sec^2(\theta) d\theta}{18 \sec(\theta)}

This identity is the key to eliminating the square root. By replacing $1 + \tan^2(\theta)$ with $\sec^2(\theta)$, we transform the denominator into a simpler form. This step highlights the power of trigonometric identities in simplifying integrals. The accurate application of these identities is crucial for the success of the trigonometric substitution method. Once the square root is eliminated, the integral becomes much more manageable.

5. Reduce the Integral

Simplifying the fraction, we get:

∫18sec⁡2(θ)dθ18sec⁡(θ)=∫sec⁡(θ)dθ\int \frac{18 \sec^2(\theta) d\theta}{18 \sec(\theta)} = \int \sec(\theta) d\theta

Here, we cancel out the common terms (18 and $\sec(\theta)$), which significantly simplifies the integral. This simplification is a direct consequence of the previous steps and brings us closer to the final solution. Reducing the integral to its simplest form is an essential part of the integration process. The resulting integral is a standard one that can be easily evaluated.

6. Integrate $\sec(\theta)\$

The integral of $\sec(\theta)$ is a standard integral, which is:

∫sec⁡(θ)dθ=ln⁡∣sec⁡(θ)+tan⁡(θ)∣+C\int \sec(\theta) d\theta = \ln|\sec(\theta) + \tan(\theta)| + C

This is a well-known result that should be memorized or easily accessible. The integral of the secant function is a fundamental result in calculus and appears frequently in various contexts. The absolute value ensures that the argument of the logarithm is positive. Adding the constant of integration $C$ is crucial for indefinite integrals, but since we are dealing with a definite integral, we will consider the limits of integration later.

7. Convert Back to $x$

We need to convert back to the original variable $x$. Since $x = 18 \tan(\theta)$, we have $\tan(\theta) = \frac{x}{18}$. To find $\sec(\theta)$, we use the identity $\sec^2(\theta) = 1 + \tan^2(\theta)$:

sec⁡(θ)=1+tan⁡2(θ)=1+(x18)2=1+x2324=324+x218\sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{x}{18}\right)^2} = \sqrt{1 + \frac{x^2}{324}} = \frac{\sqrt{324 + x^2}}{18}

This step is crucial for expressing the result in terms of the original variable. We use the initial substitution and trigonometric identities to find expressions for $\sec(\theta)$ and $\tan(\theta)$ in terms of $x$. The use of the Pythagorean identity is essential here. Accuracy in this step ensures that the final result is correct and expressed in the appropriate variable.

8. Substitute Back

Substituting $\sec(\theta)$ and $\tan(\theta)$ back into the integral, we get:

ln⁡∣324+x218+x18∣+C=ln⁡∣324+x2+x18∣+C\ln\left|\frac{\sqrt{324 + x^2}}{18} + \frac{x}{18}\right| + C = \ln\left|\frac{\sqrt{324 + x^2} + x}{18}\right| + C

This substitution expresses the integral in terms of $x$. The logarithmic form of the integral is now apparent. This step involves careful replacement of the trigonometric functions with their equivalent expressions in terms of $x$. The result is an antiderivative of the original integrand, which we can now use to evaluate the definite integral.

9. Evaluate the Definite Integral

Now, we evaluate the definite integral from 0 to 15:

∫015dx324+x2=[ln⁡∣324+x2+x18∣]015\int_0^{15} \frac{dx}{\sqrt{324 + x^2}} = \left[ \ln\left|\frac{\sqrt{324 + x^2} + x}{18}\right| \right]_0^{15}

This step involves applying the Fundamental Theorem of Calculus, which states that the definite integral of a function can be found by evaluating the antiderivative at the upper and lower limits of integration and subtracting the results. The limits of integration are crucial here, as they define the interval over which we are finding the area under the curve.

10. Apply Limits of Integration

Applying the limits of integration, we have:

ln⁡∣324+152+1518∣−ln⁡∣324+02+018∣\ln\left|\frac{\sqrt{324 + 15^2} + 15}{18}\right| - \ln\left|\frac{\sqrt{324 + 0^2} + 0}{18}\right|

We substitute the upper and lower limits into the antiderivative. This step is a straightforward application of the Fundamental Theorem of Calculus. Accuracy in this step is crucial for obtaining the correct final answer. The substitution should be done carefully, ensuring that each term is evaluated correctly.

11. Simplify the Expression

Simplify the expression:

ln⁡∣324+225+1518∣−ln⁡∣32418∣=ln⁡∣549+1518∣−ln⁡∣1818∣\ln\left|\frac{\sqrt{324 + 225} + 15}{18}\right| - \ln\left|\frac{\sqrt{324}}{18}\right| = \ln\left|\frac{\sqrt{549} + 15}{18}\right| - \ln\left|\frac{18}{18}\right|

We perform the arithmetic operations inside the logarithms. This simplification is essential for obtaining a clean and final answer. The square root and the fractions are evaluated to their simplest forms. The goal is to reduce the expression to a form that can be easily computed.

12. Final Simplification

Since $\ln(1) = 0$, the expression simplifies to:

ln⁡∣549+1518∣−ln⁡(1)=ln⁡∣549+1518∣−0=ln⁡(549+1518)\ln\left|\frac{\sqrt{549} + 15}{18}\right| - \ln(1) = \ln\left|\frac{\sqrt{549} + 15}{18}\right| - 0 = \ln\left(\frac{\sqrt{549} + 15}{18}\right)

We use the property that the natural logarithm of 1 is 0. This further simplifies the expression. The final answer is now in a compact form. This step is the culmination of all the previous steps, and it provides the numerical value of the definite integral.

13. Approximate the Value

We can approximate the value:

ln⁡(549+1518)≈ln⁡(23.43+1518)≈ln⁡(38.4318)≈ln⁡(2.135)≈0.758\ln\left(\frac{\sqrt{549} + 15}{18}\right) \approx \ln\left(\frac{23.43 + 15}{18}\right) \approx \ln\left(\frac{38.43}{18}\right) \approx \ln(2.135) \approx 0.758

This approximation provides a numerical value for the integral. The square root is approximated, and the logarithm is evaluated. This step gives a practical understanding of the magnitude of the integral. The approximate value can be useful for verifying the correctness of the solution.

Conclusion

Therefore, the value of the integral $\int_0^{15} \frac{dx}{\sqrt{324 + x^2}}$ is approximately 0.758. This evaluation demonstrates the power and utility of trigonometric substitution in solving integrals involving square roots of quadratic expressions. The step-by-step approach outlined in this article provides a clear and concise method for tackling similar problems. By understanding the underlying principles and techniques, you can confidently approach a wide range of integral calculus problems. The ability to solve such integrals is a valuable skill in mathematics and its applications in various fields.