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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.AAE.5

Evaluate the integrals in Exercises 1–6.
∫ dt / (t - √(1 - t²))

Guida verificata passo dopo passo
1
Start by examining the integral: \(\int \frac{dt}{t - \sqrt{1 - t^{2}}}\). Notice the expression in the denominator involves both \(t\) and \(\sqrt{1 - t^{2}}\), which suggests a trigonometric substitution might simplify the square root term.
Use the substitution \(t = \sin(\theta)\), which implies \(dt = \cos(\theta) d\theta\). This substitution is helpful because \(\sqrt{1 - t^{2}}\) becomes \(\sqrt{1 - \sin^{2}(\theta)} = \cos(\theta)\).
Rewrite the integral in terms of \(\theta\): replace \(t\) with \(\sin(\theta)\) and \(dt\) with \(\cos(\theta) d\theta\). The integral becomes \(\int \frac{\cos(\theta) d\theta}{\sin(\theta) - \cos(\theta)}\).
To simplify the integral, consider dividing numerator and denominator by \(\cos(\theta)\) (assuming \(\cos(\theta) \neq 0\)), which transforms the integral into \(\int \frac{d\theta}{\tan(\theta) - 1}\). This form is easier to handle.
Next, use the substitution \(u = \tan(\theta) - 1\), so that \(du = \sec^{2}(\theta) d\theta\). Express \(d\theta\) in terms of \(du\) and rewrite the integral accordingly. This will allow you to integrate with respect to \(u\) and then back-substitute to \(t\).

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Trigonometric Substitution

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