August 13, 2026 · Reference · LibreTimesIntegrals of Hyperbolic FunctionsSergeyMathematical AnalysisIndefinite integrals of the hyperbolic functions. ∫coshax dx=a1sinhax+C ∫eaxcoshbx dx=⎩⎨⎧a2−b2eax[acoshbx−bsinhbx]a=b4ae2ax+2xa=b+C ∫sinhax dx=a1coshax+C ∫eaxsinhbx dx=⎩⎨⎧a2−b2eax[−bcoshbx+asinhbx]a=b4ae2ax−2xa=b+C ∫tanhaxdx=a1lncoshax+C ∫eaxtanhbx dx=⎩⎨⎧(a+2b)e(a+2b)x2F1[1+2ba,1,2+2ba,−e2bx]−a1eax2F1[1,2ba,1+2ba,−e2bx]a=baeax−2arctan[eax]a=b+C ∫cosaxcoshbx dx=a2+b21[asinaxcoshbx+bcosaxsinhbx]+C ∫cosaxsinhbx dx=a2+b21[bcosaxcoshbx+asinaxsinhbx]+C ∫sinaxcoshbx dx=a2+b21[−acosaxcoshbx+bsinaxsinhbx]+C ∫sinaxsinhbx dx=a2+b21[bcoshbxsinax−acosaxsinhbx]+C ∫sinhaxcoshax dx=4a1cosh2ax+C ∫sinhaxcoshbx dx=a2−b21[acoshaxcoshbx−bsinhaxsinhbx]+C,a=b
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