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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 8.6.56

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.
∫ (from 0 to 1/√3) dt / (t² + 1)^(7/2)

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1
Recognize that the integral has the form \(\int \frac{dt}{(t^2 + 1)^{7/2}}\), which suggests a trigonometric substitution because of the \(t^2 + 1\) term inside the power.
Use the substitution \(t = \tan(\theta)\), which implies \(dt = \sec^2(\theta) d\theta\) and \(t^2 + 1 = \sec^2(\theta)\).
Rewrite the integral in terms of \(\theta\): replace \(dt\) with \(\sec^2(\theta) d\theta\) and \((t^2 + 1)^{7/2}\) with \((\sec^2(\theta))^{7/2} = \sec^7(\theta)\), so the integrand becomes \(\frac{\sec^2(\theta)}{\sec^7(\theta)} = \sec^{-5}(\theta)\).
Change the limits of integration from \(t\) to \(\theta\): when \(t=0\), \(\theta = \arctan(0) = 0\); when \(t=\frac{1}{\sqrt{3}}\), \(\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
Now the integral is \(\int_0^{\pi/6} \sec^{-5}(\theta) d\theta = \int_0^{\pi/6} \cos^5(\theta) d\theta\). Use a reduction formula for powers of cosine to evaluate this integral step-by-step.

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Key Concepts

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

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving expressions like √(a² + x²), √(a² - x²), or √(x² - a²). By substituting x with a trigonometric function (e.g., x = a tan θ), the integral transforms into a trigonometric integral that is often easier to evaluate.
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Reduction Formulas

Reduction formulas express an integral with a certain power or parameter in terms of a similar integral with a lower power or simpler parameter. They help solve complex integrals step-by-step by breaking them down into simpler, more manageable integrals.
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Definite Integration with Limits

Definite integration involves evaluating the integral between specified limits. When using substitution, it is important to change the limits according to the substitution or revert to the original variable before applying the limits to find the exact numerical value.
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