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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.62

62. Two integration methods Evaluate ∫ sin x cos x dx using integration by parts. Then evaluate the integral using a substitution. Reconcile your answers

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First, let's evaluate the integral \(\int \sin x \cos x \, dx\) using integration by parts. Recall the formula for integration by parts: \(\int u \, dv = uv - \int v \, du\). We need to choose \(u\) and \(dv\) from the integrand.
Choose \(u = \sin x\) and \(dv = \cos x \, dx\). Then compute \(du = \cos x \, dx\) and \(v = \int \cos x \, dx = \sin x\). Substitute these into the integration by parts formula:
\[\int \sin x \cos x \, dx = \sin x \cdot \sin x - \int \sin x \cdot \cos x \, dx = \sin^2 x - \int \sin x \cos x \, dx.\]
Notice that the integral \(\int \sin x \cos x \, dx\) appears on both sides of the equation. Let's call this integral \(I\). Then we have \(I = \sin^2 x - I\), which can be rearranged to solve for \(I\).
Next, evaluate the integral using substitution. Let \(t = \sin x\), so that \(dt = \cos x \, dx\). The integral becomes \(\int t \, dt\), which is straightforward to integrate. After integrating, substitute back \(t = \sin x\) to express the answer in terms of \(x\). Finally, compare this result with the one obtained from integration by parts to reconcile any differences, such as constant terms.

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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 based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing appropriate u and dv is crucial to simplify the integral effectively.
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Substitution Method

The substitution method simplifies an integral by changing variables to reduce it to a basic form. By setting a part of the integrand as a new variable, the integral becomes easier to evaluate. This method is especially useful when the integrand contains a function and its derivative.
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Reconciling Different Integration Results

Different integration methods can yield expressions that look different but represent the same family of functions. Reconciling these results involves using trigonometric identities or algebraic manipulation to show equivalence, often differing by a constant of integration.
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