Evaluating integrals Evaluate the following integrals.
∫√₂/₅^²/⁵ d𝓍/𝓍√(25𝓍² ―1)
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Step 1: Recognize the integral's structure. The given integral involves a square root in the denominator and a term resembling the derivative of an inverse trigonometric function. Specifically, it resembles the form of an integral involving arcsine or arctangent.
Step 2: Simplify the denominator. Notice that the denominator is √(25𝓍² - 1). Factor out 25 from the square root to rewrite it as √(25(𝓍² - 1/25)). This simplifies the expression inside the square root.
Step 3: Perform a substitution. Let u = 𝓍, and rewrite the integral in terms of u. This substitution helps simplify the integral further and prepares it for evaluation using standard formulas.
Step 4: Identify the standard formula. The integral now resembles the form ∫ du / u√(a²u² - b²), which can be solved using the formula for inverse trigonometric functions. Specifically, it matches the form of an arcsine or arctangent integral.
Step 5: Apply the formula and adjust for constants. Use the appropriate inverse trigonometric formula, ensuring to account for the constants (like 25) introduced during simplification. After applying the formula, simplify the result and include the constant of integration.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
Definite integrals represent the signed area under a curve between two specified limits. They are calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. Understanding how to evaluate definite integrals is crucial for solving problems involving areas, volumes, and other applications in calculus.
The substitution method is a technique used to simplify the process of evaluating integrals. By substituting a part of the integrand with a new variable, the integral can often be transformed into a more manageable form. This method is particularly useful when dealing with composite functions or when the integrand contains a function and its derivative.
Integrating rational functions involves finding the integral of a fraction where both the numerator and denominator are polynomials. Techniques such as partial fraction decomposition may be employed to break down complex rational functions into simpler components that are easier to integrate. Mastery of this concept is essential for evaluating integrals that appear in various calculus problems.