Textbook Question
2–9. Integrals Evaluate the following integrals.
∫₀¹ (x² / (9 − x⁶)) dx
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11. Integrals of Inverse, Exponential, & Logarithmic Functions
Integrals Involving Inverse Trigonometric Functions
4:12 minutesProblem 7.3.105
Textbook Question
Inverse identity Show that cosh⁻¹(cosh x) = |x| by using the formula cosh⁻¹ t = ln (t + √(t² – 1)) and considering the cases x ≥ 0 and x < 0.
Verified step by step guidance1
Recall the definition of the inverse hyperbolic cosine function: \(\cosh^{-1} t = \ln \left(t + \sqrt{t^2 - 1} \right)\) for \(t \geq 1\).
Substitute \(t = \cosh x\) into the formula to get \(\cosh^{-1}(\cosh x) = \ln \left( \cosh x + \sqrt{\cosh^2 x - 1} \right)\).
Use the identity \(\cosh^2 x - 1 = \sinh^2 x\) to rewrite the expression as \(\ln \left( \cosh x + |\sinh x| \right)\).
Consider the two cases separately:
- For \(x \geq 0\), \(\sinh x \geq 0\), so \(|\sinh x| = \sinh x\).
- For \(x < 0\), \(\sinh x < 0\), so \(|\sinh x| = -\sinh x\).
Evaluate the expression in each case:
- When \(x \geq 0\), \(\ln (\cosh x + \sinh x) = \ln (e^x) = x\).
- When \(x < 0\), \(\ln (\cosh x - \sinh x) = \ln (e^{-x}) = -x = |x|\).
Thus, \(\cosh^{-1}(\cosh x) = |x|\).
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4mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Hyperbolic Cosine Function
The inverse hyperbolic cosine, denoted as cosh⁻¹(t), is the function that returns the value x such that cosh(x) = t for t ≥ 1. It is defined by the formula cosh⁻¹(t) = ln(t + √(t² – 1)), which expresses the inverse in terms of natural logarithms and square roots.
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Properties of the Hyperbolic Cosine Function
The hyperbolic cosine function, cosh(x), is an even function, meaning cosh(x) = cosh(–x). It is always greater than or equal to 1 and increases exponentially for large |x|. This symmetry is key to understanding why cosh⁻¹(cosh x) equals the absolute value of x.
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Piecewise Analysis Based on Domain
To prove cosh⁻¹(cosh x) = |x|, it is essential to consider two cases: x ≥ 0 and x < 0. Since cosh is even, analyzing these cases separately helps show that the inverse function returns x when x is nonnegative and –x when x is negative, resulting in the absolute value.
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