Edexcel IAL - Further Pure Mathematics 3- 4.1 Integration of Hyperbolic Functions- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 3- 4.1 Integration of Hyperbolic Functions -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 3- 4.1 Integration of Hyperbolic Functions -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.1 Integration of Hyperbolic Functions
Integration of Hyperbolic Functions
Hyperbolic functions can be integrated using standard results, closely mirroring the integration of trigonometric functions. More complicated expressions may require substitution, algebraic manipulation, or the use of known hyperbolic identities.
Standard Integrals of Hyperbolic Functions
| Function | Integral |
|---|---|
| \( \sinh x \) | \( \cosh x + C \) |
| \( \cosh x \) | \( \sinh x + C \) |
| \( \mathrm{ sech^2 x} \) | \( \tanh x + C \) |
| \( \mathrm{csch^2 x} \) | \( -\coth x + C \) |
| \( \mathrm{sech x\,\tanh x} \) | \( -\mathrm{sech x + C} \) |
| \( \mathrm{ csch x\,\coth x }\) | \( – \mathrm{csch x + C} \) |
Useful Hyperbolic Identities for Integration
- \( \cosh^2 x – \mathrm{sinh^2 x = 1 }\)
- \( 1 – \tanh^2 x = \mathrm{sech^2 x }\)
- \( \coth^2 x – 1 = \mathrm{csch^2 x} \)
These identities are often used to rewrite an integrand into a standard form.
Integration Techniques
- Use substitution when the argument is a function of \( x \)
- Rewrite powers using identities where possible
- Check whether the integrand matches a known derivative
Example :
Evaluate
\( \displaystyle \int \cosh 3x \, dx \).
▶️ Answer/Explanation
Use substitution.
Let \( u = 3x \), so \( du = 3\,dx \)
\( \int \cosh 3x \, dx = \dfrac{1}{3} \int \cosh u \, du \)
\( = \dfrac{1}{3}\sinh u + C = \dfrac{1}{3}\sinh 3x + C \)
Conclusion: \( \dfrac{1}{3}\sinh 3x + C \).
Example :
Evaluate
\( \displaystyle \int x\,\mathrm{sech^2 x} \, dx \).
▶️ Answer/Explanation
Use integration by parts.
Let \( u = x \), \( dv = \mathrm{sech^2 x}\,dx \)
Then \( du = dx \), \( v = \tanh x \)
\( \int x\mathrm{sech^2 x}\,dx = x\tanh x – \int \tanh x\,dx \)
\( = x\tanh x – \ln(\cosh x) + C \)
Conclusion: \( x\tanh x – \ln(\cosh x) + C \).
Example :
Evaluate
\( \displaystyle \int \sinh^2 x \, dx \).
▶️ Answer/Explanation
Use the identity
\( \sinh^2 x = \cosh^2 x – 1 \)
\( \int \sinh^2 x\,dx = \int (\cosh^2 x – 1)\,dx \)
Since \( \cosh^2 x = \dfrac{1+\cosh 2x}{2} \):
\( = \dfrac{1}{2}\int \cosh 2x\,dx – \dfrac{1}{2}\int dx \)
\( = \dfrac{1}{4}\sinh 2x – \dfrac{x}{2} + C \)
Conclusion: \( \dfrac{1}{4}\sinh 2x – \dfrac{x}{2} + C \).