Evaluate The Following Integral Using Trigonometric Substitution.

Evaluating integrals using trigonometric substitution is a powerful technique that allows us to simplify complex integrals involving square roots of quadratic expressions. This method leverages trigonometric identities to transform the integral into a more manageable form. Trigonometric substitution is particularly useful when dealing with integrals containing expressions like a² - x², a² + x², or x² - a².

Understanding Trigonometric Substitution

The core idea behind trigonometric substitution is to replace the variable of integration, typically x, with a trigonometric function. This substitution aims to eliminate the square root in the integral, thereby simplifying the expression. The choice of the appropriate trigonometric function depends on the form of the expression under the square root:

• For integrals containing √(a² - x²): Use the substitution x = a sin θ.
• For integrals containing √(a² + x²): Use the substitution x = a tan θ.
• For integrals containing √(x² - a²): Use the substitution x = a sec θ.

After the substitution, the integral is transformed into an integral involving trigonometric functions, which can often be solved using standard integration techniques and trigonometric identities.

Steps Involved in Trigonometric Substitution

The process of evaluating integrals using trigonometric substitution generally involves the following steps:

1. Identify the Appropriate Substitution: Determine which trigonometric substitution is suitable based on the form of the expression under the square root.
2. Perform the Substitution: Replace x with the chosen trigonometric function and compute dx in terms of dθ.
3. Simplify the Integral: Substitute x and dx into the original integral and simplify the expression using trigonometric identities.
4. Evaluate the Trigonometric Integral: Solve the resulting trigonometric integral using standard integration techniques.
5. Convert Back to the Original Variable: Replace the trigonometric functions with expressions involving x using the original substitution to obtain the final answer.

Case 1: Integrals Involving √(a² - x²)

Let's consider integrals of the form ∫f(x) dx, where f(x) involves √(a² - x²). The appropriate substitution for this case is:

• x = a sin θ
• dx = a cos θ dθ

Using the identity sin² θ + cos² θ = 1, we can simplify the expression under the square root as follows:

√(a² - x²) = √(a² - (a sin θ)²) = √(a² - a² sin² θ) = √(a²(1 - sin² θ)) = √(a² cos² θ) = a cos θ

Example: Evaluate the integral ∫√(4 - x²) dx.

1. Identify the Substitution: Here, a² = 4, so a = 2. We use the substitution x = 2 sin θ.
2. Perform the Substitution:
   • x = 2 sin θ
   • dx = 2 cos θ dθ
3. Simplify the Integral: ∫√(4 - x²) dx = ∫√(4 - (2 sin θ)²) (2 cos θ dθ) = ∫√(4 - 4 sin² θ) (2 cos θ dθ) = ∫√(4(1 - sin² θ)) (2 cos θ dθ) = ∫2 cos θ (2 cos θ dθ) = 4∫cos² θ dθ
4. Evaluate the Trigonometric Integral: Using the identity cos² θ = (1 + cos 2θ) / 2, we get: 4∫cos² θ dθ = 4∫(1 + cos 2θ) / 2 dθ = 2∫(1 + cos 2θ) dθ = 2(θ + (1/2)sin 2θ) + C = 2θ + sin 2θ + C
5. Convert Back to the Original Variable: Since x = 2 sin θ, we have sin θ = x/2, so θ = arcsin(x/2). Also, sin 2θ = 2 sin θ cos θ. We know sin θ = x/2, and cos θ = √(1 - sin² θ) = √(1 - (x/2)²) = √(4 - x²) / 2. Thus, sin 2θ = 2 (x/2) (√(4 - x²) / 2) = x√(4 - x²) / 2. Therefore, the final answer is: 2θ + sin 2θ + C = 2 arcsin(x/2) + (x√(4 - x²) / 2) + C

Case 2: Integrals Involving √(a² + x²)

For integrals of the form ∫f(x) dx, where f(x) involves √(a² + x²), the appropriate substitution is:

• x = a tan θ
• dx = a sec² θ dθ

Using the identity 1 + tan² θ = sec² θ, we can simplify the expression under the square root as follows:

√(a² + x²) = √(a² + (a tan θ)²) = √(a² + a² tan² θ) = √(a²(1 + tan² θ)) = √(a² sec² θ) = a sec θ

Example: Evaluate the integral ∫1 / √(9 + x²) dx.

1. Identify the Substitution: Here, a² = 9, so a = 3. We use the substitution x = 3 tan θ.
2. Perform the Substitution:
   • x = 3 tan θ
   • dx = 3 sec² θ dθ
3. Simplify the Integral: ∫1 / √(9 + x²) dx = ∫1 / √(9 + (3 tan θ)²) (3 sec² θ dθ) = ∫1 / √(9 + 9 tan² θ) (3 sec² θ dθ) = ∫1 / √(9(1 + tan² θ)) (3 sec² θ dθ) = ∫1 / (3 sec θ) (3 sec² θ dθ) = ∫sec θ dθ
4. Evaluate the Trigonometric Integral: ∫sec θ dθ = ln |sec θ + tan θ| + C
5. Convert Back to the Original Variable: Since x = 3 tan θ, we have tan θ = x/3. We also have sec θ = √(1 + tan² θ) = √(1 + (x/3)²) = √(9 + x²) / 3. Therefore, the final answer is: ln |sec θ + tan θ| + C = ln |(√(9 + x²) / 3) + (x / 3)| + C = ln |√(9 + x²) + x| - ln 3 + C Since ln 3 is a constant, we can absorb it into the constant of integration: ln |√(9 + x²) + x| + C

Case 3: Integrals Involving √(x² - a²)

For integrals of the form ∫f(x) dx, where f(x) involves √(x² - a²), the appropriate substitution is:

• x = a sec θ
• dx = a sec θ tan θ dθ

Using the identity sec² θ - 1 = tan² θ, we can simplify the expression under the square root as follows:

√(x² - a²) = √((a sec θ)² - a²) = √(a² sec² θ - a²) = √(a²(sec² θ - 1)) = √(a² tan² θ) = a tan θ

Example: Evaluate the integral ∫√(x² - 16) / x dx.

1. Identify the Substitution: Here, a² = 16, so a = 4. We use the substitution x = 4 sec θ.
2. Perform the Substitution:
   • x = 4 sec θ
   • dx = 4 sec θ tan θ dθ
3. Simplify the Integral: ∫√(x² - 16) / x dx = ∫√((4 sec θ)² - 16) / (4 sec θ) (4 sec θ tan θ dθ) = ∫√(16 sec² θ - 16) / (4 sec θ) (4 sec θ tan θ dθ) = ∫√(16(sec² θ - 1)) / (4 sec θ) (4 sec θ tan θ dθ) = ∫4 tan θ / (4 sec θ) (4 sec θ tan θ dθ) = ∫4 tan² θ dθ
4. Evaluate the Trigonometric Integral: Using the identity tan² θ = sec² θ - 1, we get: 4∫tan² θ dθ = 4∫(sec² θ - 1) dθ = 4(tan θ - θ) + C
5. Convert Back to the Original Variable: Since x = 4 sec θ, we have sec θ = x/4, so θ = arcsec(x/4). We also have tan θ = √(sec² θ - 1) = √((x/4)² - 1) = √(x² - 16) / 4. Therefore, the final answer is: 4(tan θ - θ) + C = 4((√(x² - 16) / 4) - arcsec(x/4)) + C = √(x² - 16) - 4 arcsec(x/4) + C

Additional Considerations and Tips

• Choosing the Correct Substitution: The key to successful trigonometric substitution lies in selecting the appropriate substitution based on the form of the integral. Always identify the expression under the square root and match it with one of the three cases discussed above.
• Simplifying Trigonometric Expressions: After the substitution, the integral will involve trigonometric functions. Use trigonometric identities to simplify the expression before attempting to integrate.
• Drawing Reference Triangles: When converting back to the original variable, drawing a right triangle can be helpful. Use the original substitution to label the sides of the triangle and then find the other trigonometric functions in terms of x.
• Dealing with Definite Integrals: If you are evaluating a definite integral using trigonometric substitution, you can either convert the limits of integration to values of θ or convert back to the original variable x before applying the original limits.
• Completing the Square: Sometimes, the expression under the square root is not in the standard form (a² - x², a² + x², or x² - a²). In such cases, you may need to complete the square to bring the expression into the required form before applying trigonometric substitution.

Example: Completing the Square:

Consider the integral ∫1 / √(8 + 2x - x²) dx.

1. Complete the Square: Rewrite the expression under the square root: 8 + 2x - x² = 8 - (x² - 2x) = 8 - (x² - 2x + 1 - 1) = 8 - ((x - 1)² - 1) = 9 - (x - 1)²
2. Identify the Substitution: Now the integral becomes ∫1 / √(9 - (x - 1)²) dx. This is in the form √(a² - u²), where a = 3 and u = x - 1. We use the substitution x - 1 = 3 sin θ, so x = 1 + 3 sin θ and dx = 3 cos θ dθ.
3. Perform the Substitution and Simplify: ∫1 / √(9 - (x - 1)²) dx = ∫1 / √(9 - (3 sin θ)²) (3 cos θ dθ) = ∫1 / √(9 - 9 sin² θ) (3 cos θ dθ) = ∫1 / √(9(1 - sin² θ)) (3 cos θ dθ) = ∫1 / (3 cos θ) (3 cos θ dθ) = ∫dθ
4. Evaluate the Trigonometric Integral: ∫dθ = θ + C
5. Convert Back to the Original Variable: Since x - 1 = 3 sin θ, we have sin θ = (x - 1) / 3, so θ = arcsin((x - 1) / 3). Therefore, the final answer is: θ + C = arcsin((x - 1) / 3) + C

Advanced Examples and Techniques

Let's explore some more complex examples that require a deeper understanding of trigonometric substitution and integration techniques.

Example 1: Evaluate the integral ∫x³ / √(x² + 4) dx.

1. Identify the Substitution: We have √(x² + 4), so we use the substitution x = 2 tan θ, and dx = 2 sec² θ dθ.
2. Perform the Substitution and Simplify: ∫x³ / √(x² + 4) dx = ∫(2 tan θ)³ / √((2 tan θ)² + 4) (2 sec² θ dθ) = ∫8 tan³ θ / √(4 tan² θ + 4) (2 sec² θ dθ) = ∫8 tan³ θ / √(4(tan² θ + 1)) (2 sec² θ dθ) = ∫8 tan³ θ / (2 sec θ) (2 sec² θ dθ) = 8∫tan³ θ sec θ dθ
3. Evaluate the Trigonometric Integral: Rewrite the integral: 8∫tan³ θ sec θ dθ = 8∫tan² θ (tan θ sec θ) dθ = 8∫(sec² θ - 1) (tan θ sec θ) dθ Now, let u = sec θ, so du = tan θ sec θ dθ: 8∫(u² - 1) du = 8(u³/3 - u) + C = 8(sec³ θ / 3 - sec θ) + C
4. Convert Back to the Original Variable: Since x = 2 tan θ, we have tan θ = x/2. We also have sec θ = √(1 + tan² θ) = √(1 + (x/2)²) = √(4 + x²) / 2. Therefore, the final answer is: 8(sec³ θ / 3 - sec θ) + C = 8((√(4 + x²) / 2)³ / 3 - (√(4 + x²) / 2)) + C = (√(4 + x³)) / 3 - 4√(4 + x²) + C

Example 2: Evaluate the integral ∫1 / (x²√(x² - 1)) dx.

1. Identify the Substitution: We have √(x² - 1), so we use the substitution x = sec θ, and dx = sec θ tan θ dθ.
2. Perform the Substitution and Simplify: ∫1 / (x²√(x² - 1)) dx = ∫1 / (sec² θ √(sec² θ - 1)) (sec θ tan θ dθ) = ∫1 / (sec² θ tan θ) (sec θ tan θ dθ) = ∫1 / sec θ dθ = ∫cos θ dθ
3. Evaluate the Trigonometric Integral: ∫cos θ dθ = sin θ + C
4. Convert Back to the Original Variable: Since x = sec θ, we have sec θ = x, so θ = arcsec(x). We also have sin θ = √(1 - cos² θ) = √(1 - (1/sec² θ)) = √(1 - (1/x²)) = √(x² - 1) / x. Therefore, the final answer is: sin θ + C = √(x² - 1) / x + C

Practical Applications

Trigonometric substitution is not just a theoretical exercise; it has practical applications in various fields, including:

• Physics: Calculating the arc length of curves, finding the center of mass of objects with certain shapes, and solving problems in mechanics involving circular motion.
• Engineering: Designing structures, analyzing stress and strain in materials, and calculating areas and volumes of complex shapes.
• Computer Graphics: Creating realistic images and animations by modeling curved surfaces and calculating lighting effects.

Conclusion

Trigonometric substitution is a versatile and essential technique in integral calculus. By understanding the different cases and mastering the steps involved, you can tackle a wide range of integrals that would otherwise be difficult or impossible to solve. Remember to choose the appropriate substitution, simplify the integral using trigonometric identities, and convert back to the original variable to obtain the final answer. With practice and patience, you can become proficient in using trigonometric substitution to solve even the most challenging integration problems.