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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 8.2.32

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ (cos(√x))/(√x) dx

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Start by identifying a suitable substitution to simplify the integral. Notice that the integrand contains both \( \cos(\sqrt{x}) \) and \( \frac{1}{\sqrt{x}} \), which suggests substituting \( u = \sqrt{x} \).
Express \( u = \sqrt{x} \) in terms of \( x \), so \( u = x^{1/2} \). Then, differentiate both sides with respect to \( x \) to find \( du \): \( du = \frac{1}{2} x^{-1/2} dx = \frac{1}{2\sqrt{x}} dx \).
Solve for \( dx \) in terms of \( du \) and \( x \): \( dx = 2 \sqrt{x} du \). Substitute \( dx \) and \( \sqrt{x} = u \) back into the integral to rewrite it entirely in terms of \( u \).
After substitution, the integral becomes \( \int \frac{\cos(u)}{u} \cdot 2u \, du \). Simplify the expression inside the integral by canceling terms where possible.
Integrate the simplified integral with respect to \( u \), then substitute back \( u = \sqrt{x} \) to express the answer in terms of \( x \).

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Key Concepts

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

Substitution Method

The substitution method simplifies integrals by changing variables to transform the integral into a more familiar form. For example, setting a new variable equal to a function inside the integral can make the integral easier to evaluate.
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Integration of Trigonometric Functions

Understanding how to integrate trigonometric functions like cosine is essential. The integral of cos(u) with respect to u is sin(u) plus a constant, which helps in solving integrals involving trigonometric expressions.
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Handling Radicals in Integrals

Integrals involving radicals, such as square roots, often require rewriting the expression in terms of powers or using substitution to simplify the integral. Recognizing how to manipulate radicals is key to evaluating such integrals.
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