3.4 The Hyperbolic Substitution
In general the following will give a guide
- 1.
- If the integrand involves the expression \(a^2 + x^2\), substitute \(x = a\sinh \theta \)
- 2.
- If the integrand involves the expression \(a^2 - x^2\), substitute \(x = a\tanh \theta \)
- 3.
- If the integrand involves the expression \(x^2 - a^2\), substitute \(x = a\cosh \theta \)
For the types of substitutions the most important identities to use are
- 1.
- \(\cosh ^2\theta - \sinh ^2\theta = 1\)
- 2.
- \(2\sinh \theta \cosh \theta = \sinh 2 \theta \)
- 3.
- \(\cosh ^2\theta = \dfrac {1}{2}\big (1 + \cosh 2 \theta \big )\)
- 4.
- \(\sinh ^2\theta = \dfrac {1}{2}\big ( \cosh 2 \theta - 1\big )\)
Solution.
Let \(\hspace {0.2cm} x = \sinh \theta \hspace {0.2cm}\). Then \(\hspace {0.2cm} dx = \cosh \theta \hspace {0.1cm}d\theta \). Thus
\begin {align*} \int \sqrt {x^2 + 1}\hspace {0.2cm}dx & = \int \sqrt {\sinh ^2 \theta + 1}\cdot \cosh \theta \hspace {0.2cm} d\theta \\\\ & = \int \sqrt {\cosh ^2\theta }\cdot \cosh \theta \hspace {0.2cm}d\theta \\\\ & = \int \cosh ^2 \theta \hspace {0.2cm}d\theta \\\\ & = \dfrac {1}{2}\int \big (1 + \cosh 2 \theta \big ) \hspace {0.2cm}d\theta \\\\ & = \dfrac {1}{2}\Big [ \theta + \dfrac {1}{2} \sinh 2\theta \Big ] + c\\\\ & = \dfrac {1}{2}\Big [\sinh ^{-1}x + x\sqrt {1 + x^2}\Big ] + c\\\\ \end {align*}
Find \(\hspace {0.3cm} \displaystyle {\int \dfrac {1}{x\sqrt {1 - x^2}}\hspace {0.2cm}dx}\)
Solution.
Let \(\hspace {0.2cm} x = \sec h\theta \hspace {0.2cm}\). Then \(\hspace {0.2cm} dx = -\sec h\theta \tanh \theta \hspace {0.2cm}d\theta \)
Now, \(\hspace {0.2cm} 1 - x^2 = 1 - \sec h^2\theta = \tanh ^2\theta \)
\begin {align*} \implies \hspace {0.5cm} \int \dfrac {1}{x\sqrt {1 - x^2}}\hspace {0.2cm}dx & = - \int \dfrac {\sec h\theta \tanh \theta \hspace {0.2cm}d\theta }{\sec h\theta \tanh \theta }\\\\ & = -\int d\theta \\\\ & = -\theta + c\\ & = -\sec h^{-1} x + c\\\\\\ \end {align*}
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