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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.PE.4

Evaluate the integrals in Exercises 1–8 using integration by parts.
∫ arccos(x / 2) dx

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1
Identify the integral to solve: \(\int \arccos\left(\frac{x}{2}\right) \, dx\).
Choose parts for integration by parts: let \(u = \arccos\left(\frac{x}{2}\right)\) and \(dv = dx\).
Compute \(du\): differentiate \(u\) with respect to \(x\). Use the chain rule on \(\arccos\left(\frac{x}{2}\right)\), recalling that \(\frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}}\). So, \(du = -\frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^2}} \cdot \frac{1}{2} \, dx = -\frac{1}{2 \sqrt{1 - \frac{x^2}{4}}} \, dx\).
Compute \(v\): integrate \(dv = dx\), so \(v = x\).
Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substitute the expressions for \(u\), \(v\), and \(du\) to set up the new integral to solve.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is crucial to simplify the integral effectively.
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Inverse Trigonometric Functions

Inverse trigonometric functions, like arccos(x), are the inverses of trigonometric functions and have specific derivatives. For arccos(x), the derivative is -1/√(1 - x²). Understanding these derivatives helps in differentiating parts of the integrand during integration by parts.
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Substitution and Simplification

Substitution involves changing variables to simplify integrals, especially when dealing with composite functions like arccos(x/2). Simplifying the resulting expressions after applying integration by parts often requires algebraic manipulation or trigonometric identities to reach a solvable form.
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