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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.5.116
Chapter 5, Problem 5.5.116

Multiple substitutions If necessary, use two or more substitutions to find the following integrals.                                                                                    
                                                                                                                                                                    
  ∫ 𝓍 sin⁴ 𝓍² cos 𝓍² d𝓍 (Hint: Begin with u = 𝓍², and then use v = sin u .)

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1
Start by identifying the first substitution as suggested: let \(u = x^{2}\). Then, compute the differential \(du = 2x \, dx\), which implies \(x \, dx = \frac{du}{2}\).
Rewrite the integral in terms of \(u\). Notice that \(\sin^{4}(x^{2}) = \sin^{4}(u)\) and \(\cos(x^{2}) = \cos(u)\). The integral becomes \(\int x \sin^{4}(x^{2}) \cos(x^{2}) \, dx = \int \sin^{4}(u) \cos(u) \cdot x \, dx = \int \sin^{4}(u) \cos(u) \cdot \frac{du}{2}\).
Simplify the integral to \(\frac{1}{2} \int \sin^{4}(u) \cos(u) \, du\). Now, observe that the integrand involves \(\sin^{4}(u)\) and \(\cos(u)\), which suggests a second substitution.
For the second substitution, let \(v = \sin(u)\). Then, \(dv = \cos(u) \, du\). This transforms the integral into \(\frac{1}{2} \int v^{4} \, dv\).
Now, the integral is a straightforward power integral in terms of \(v\). Integrate \(\frac{1}{2} \int v^{4} \, dv\) by increasing the power by one and dividing by the new exponent, then substitute back \(v = \sin(u)\) and \(u = x^{2}\) 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 in Integration

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It involves choosing a substitution u = g(x) such that the integral in terms of u is easier to evaluate. This technique is especially useful when the integral contains composite functions or products of functions.
Recommended video:
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Euler's Method

Multiple Substitutions

Multiple substitutions involve applying the substitution method more than once to solve complex integrals. After the first substitution simplifies part of the integral, a second substitution can further reduce it. This stepwise approach is essential when a single substitution does not fully simplify the integral.
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Substitution With an Extra Variable

Trigonometric Functions and Their Derivatives

Understanding the derivatives and integrals of trigonometric functions like sine and cosine is crucial. For example, knowing that the derivative of sin u is cos u helps in choosing substitutions and simplifying integrals involving powers of sine and cosine. This knowledge aids in recognizing patterns and applying substitutions effectively.
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Derivatives of Other Inverse Trigonometric Functions
Related Practice
Textbook Question

Integrals with sin² 𝓍 and cos² 𝓍 Evaluate the following integrals.                                                                                                             

                                                                                                                                                                    

 ∫ sin² 𝓍 d𝓍

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Textbook Question

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Textbook Question

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ƒ(𝓍) = 1/(𝓍² + 1) on [―1, 1]

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Textbook Question

Areas of regions Find the area of the following regions.                                                                                                                   

                                                                                                                                                                 The region bounded by the graph of ƒ(𝓍) = (𝓍―4)⁴ and the 𝓍-axis between and 𝓍 = 2 and 𝓍= 6

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Textbook Question

Integrals with sin² 𝓍 and cos² 𝓍 Evaluate the following integrals.                                                                                                             

                                                                                                                                                                    

 ∫₀^π/⁴ cos² 8θ dθ

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