Evaluating an Indefinite Integral Using Trigonometric Substitution

Introduction

• Goal: Evaluate an indefinite integral with a square root in the denominator.
• Noticing a pattern: Square root resembles a Pythagorean identity (used in right triangles).
• Insight: Recognize square root of 4 - x^2 as a non-hypotenuse side in a right triangle.

Setting up the Problem

• Consider the hypotenuse of length 2 and one side of length x in a right triangle.
• Use Pythagorean theorem:
  • Other side length = sqrt(4 - x^2)

Trigonometric Substitution

• Triangle setup:
  • Hypotenuse = 2
  • One side = x
  • Other side = sqrt(4 - x^2)
• Using theta as an angle in the triangle:
  • Sine function: sin(theta) = x/2 -> x = 2 * sin(theta)
  • Cosine function: cos(theta) = sqrt(4 - x^2) / 2
    • Therefore, sqrt(4 - x^2) = 2 * cos(theta)

Performing Substitution in the Integral

• Substituting x = 2 * sin(theta):
  • dx = 2 * cos(theta) dtheta
• Substituting into the integral:
  • Numerator: dx = 2 * cos(theta) dtheta
  • Denominator: sqrt(4 - x^2) = 2 * cos(theta)
• Integral simplifies to:
  • Integral of (1 dtheta) = theta + C

Solving for theta

• x = 2 * sin(theta)
  • sin(theta) = x/2
  • theta = arcsin(x/2)
• Final result for the integral:
  • Integral = arcsin(x/2) + C

Domain Considerations

• Original function domain: -2 < x < 2
• Ensuring substitution preserves domain:
  • -1 < sin(theta) < 1 corresponds to -pi/2 < theta < pi/2
• Ensure cosine of theta does not equal zero in the domain.

Conclusion

• The trigonometric substitution is valid within the specified domain.
• Result is consistent with trigonometric identity restrictions.
• Integral successfully evaluated to arcsin(x/2) + C without domain issues.