Mathematics Dictionary Dr. K. G. Shih

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Integral of hyperbolic functions

Subjects

Symbol Defintion Example : Sqr(x) is square root of x Q01 | - ∫sinh(x)dx = cosh(x) + C Q02 | - ∫cosh(x)dx = sinh(x) + C Q03 | - ∫tanh(x)dx = ln(cosh(x)) + C Q04 | - ∫csch(x)dx = ln(csch(x)) - coth(x)) + C Q05 | - ∫sech(x)dx = ln(sech(x) + tanh(x)) + C Q06 | - ∫coth(x)dx = + C Q07 | - ∫(sinh(x)^2)dx = ? Q08 | - ∫(cosh(x)^2)dx = ? Q09 | - ∫(tanh(x)^2)dx = ? Q10 | - ∫(csch(x)^2)dx = ? Q11 | - ∫(sech(x)^2)dx = ? Q12 | - ∫(coth(x)^2)dx = ? Q13 | - ∫(sinh(x)*cosh(x))dx = ? Q14 | - ∫(sinh(2*x))dx = ?

Answers

Q01. ∫sinh(x)dx = cos(x) + C Anti-derivative Since we know that d/dx(cosh(x)) = sinh(x) Hence d(cosh(x)) = sinh(x)dx Hence cosh(x) = ∫sinh(x)dx Hence ∫sinh(x)dx = cosh(x) + C 2nd method ∫cosh(x)dx = ∫((exp^x + exp^(-x))/2)dx = (exp^x - exp^(-x))/2 = sinh(x) + C Go to Begin Q02. ∫ cosh(x)dx = sinh(x) + C Anti-derivative Since we know that d/dx(sinh(x)) = cosh(x) Hence d(sinh(x)) = cosh(x)dx Hence sinh(x) = ∫cosh(x)dx Hence ∫cosh(x)dx = sinh(x) + C 2nd method ∫cosh(x)dx = ∫((exp^x + exp^(-x))/2)dx = (exp^x - exp^(-x))/2 = sinh(x) + C Go to Begin Q03. ∫ tan(x)dx = ln(cosh(x)) + C There is no anti-derivative Since we know that d/dx(tan(x)) = sec(x)^2 Hence d(tan(x)) = (sec(x)^2)dx Hence tan(x) = ∫(sec(x)^2)dx Hence ∫(sec(x)^2)dx = tan(x) + C We can not get intgration of tan(x) by anti-derivative Porve that integration of tan(x) is -ln(cos(x)) tanh(x)dx = (sinh(x)dx)/cosh(x) sinh(x)dx = d(cosh(x)) Let cosh(x) = u, then du = sinh(x)dx Hence tanh(x)dx = du/u Hence ∫tanh(x)dx = ln(u) + C = ln(cosh(x)) + C Verify d/dx(ln(cosh(x)) + C) = (1/cosh(x))*(d/dx(cosh(x))) = sinh(x)/cosh(x) = tanh(x) Go to Begin Q04. ∫ csch(x)dx = ln(csch(x)) - coth(x)) It is hard to work from left to right. We will find d/dx(F(x)) = csch(x) From right side, we prove d/dx(ln(csch(x) - coth(x)) = csch(x) use chain rule and d/dx(ln(u)) = (1/u)*(du/dx) d/dx(ln(csch(x) - coth(x)) = (1/(csch(x) - coth(x))*(d/dx(csch(x) - coth(x)) = (1/(csch(x) - coth(x))*(csch(x)*coth(x) - (csch(x)^2) = (1/(csch(x) - coth(x))*(csch(x)*(coth(x) - csch(x)) = csch(x) Hence csch(x)dx = d(ln(csch(x) - coth(x)) Hence ∫ csch(x)dx = ln(csch(x) - coth(x)) Go to Begin Q05. ∫ sech(x)dx = ln(sech(x)) + tanh(x)) It is hard to work from left to right. We will find d/dx(F(x)) = sech(x) From right side, we prove d/dx(ln(sec(x) + tan(x)) = sec(x) use chain rule and d/dx(ln(u)) = (1/u)*(du/dx) d/dx(ln(sech(x) + tanh(x)) = (1/(sech(x) + tanh(x))*(d/dx(sech(x) + tanh(x)) = (1/(sech(x) + tanh(x))*(sech(x)*tanh(x) + sech(x)^2) = (1/(sech(x) + tanh(x))*(sech(x)*(tanh(x) + sech(x)) = sech(x) Hence sech(x)dx = d(ln(sech(x) + tanh(x)) Hence ∫ sech(x)dx = ln(sech(x) + tanh(x)) Go to Begin Q06. ∫ coth(x)dx = +ln(sinh(x)) + C There is no anti-derivative Since we know that d/dx(coth(x)) = csch(x)^2 Hence d(coth(x)) = (ssch(x)^2)dx Hence coth(x) = ∫(csch(x)^2)dx Hence ∫(csch(x)^2)dx = coth(x) + C We can not get intgration of coth(x) by anti-derivative Porve that integration of coth(x) is ln(sinh(x)) coth(x)dx = (cosh(x)dx)/sinh(x) cosh(x)dx = d(sinh(x))dx Let sinh(x) = u and du = cosh(x)dx Then coth(x)dx = du/u Hence ∫coth(x)dx = ln(u) + C = ln(sinh(x)) + C Go to Begin Q07. ∫(sinh(x)^2)dx = ? No anti-derivative is given Use cosh(2*x) = 1 + 2*sinh(x)^2 Hence sinh(x)^2 = (1 + cosh(2*x))/2 ∫(sinh(x)^2)dx = ∫((1 + cosh(2*x)/2)dx = x/2 + sinh(2*x)/4 Verify d/dx(x/2 + sinh(2*x)/4) = 1/2 + (1/2)*cosh(2*x) = (1 + cosh(2*x))/2 = sinh(x)^2 Go to Begin Q08. ∫(cosh(x)^2)dx = ? No anti-derivative is given Use cosh(2*x) = 2*cosh(x)^2 - 1 Hence cosh(x)^2 = (1 + cosh(2*x))/2 ∫(cosh(x)^2)dx = ∫((1 + cosh(2*x)/2)dx = x/2 + sinh(2*x)/4 Verify d/dx(x/2 + sinh(2*x)/4) = 1/2 + (1/2)*cosh(2*x) = (1 + cosh(2*x))/2 = cosh(x)^2 Go to Begin Q09. ∫(tanh(x)^2)dx = ? No anti-derivative is given Use 1 + tanh(x)^2 = sech(x)^2 and d/dx(tanh(x)) = sech(x)^2 Hence tanh(x)^2 = sech(2*x)^2 - 1 ∫(tanh(x)^2)dx = ∫((sech(x)^2 - 1)dx = tanh(x) - x Verify d/dx(tanh(x) - x) = sech(x)^2 - 1 = tanh(x)^2 Go to Begin Q10. ∫(csch(x)^2)dx = ? There is anti-derivative. Since d/dx(coth(x)) = csch(x)^2 Hence d(coth(x)) = (csch(2*x)^2)dx ∫(csch(x)^2)dx = coth(x) + C Verify d/dx(coth(x)) = csch(x)^2 Go to Begin Q11. ∫(sech(x)^2)dx = ? There is anti-derivative. Since d/dx(tanh(x)) = sech(x)^2 Hence d(tanh(x)) = (sech(2*x)^2)dx ∫(sech(x)^2)dx = tanh(x) + C Verify d/dx(tanh(x)) = sech(x)^2 Go to Begin Q12. ∫(coth(x)^2)dx = ? No anti-derivative is given Use 1 + coth(x)^2 = csch(x)^2 and d/dx(coth(x)) = csch(x)^2 Hence coth(x)^2 = csch(x)^2 - 1 ∫(coth(x)^2)dx = ∫((csch(x)^2 - 1)dx = coth(x) - x Verify d/dx(coth(x) - x) = csch(x)^2 - 1 = coth(x)^2 Go to Begin Q13. ∫(sinh(x)*cosh(x))dx = ? No anti-derivative is given Let u = sinh(x) and then du = cosh(x)dx Hence the integrant (sinh(x)*cosh(x))dx = udu ∫(sinh(x)*cosh(x))dx = ∫udu = u^2/2 = (sinh(x)^2)/2 + C Verify d/dx((sinh(x)^2/)/2) = 2*sinh(x)*cosh(x)/2 = sinh(x)*cosh(x) Second method : ∫(sin(x)*cos(x))dx = ? Let u = cosh(x) and then du = sinh(x)dx Hence the integrant (sinh(x)*cosh(x))dx = -udu ∫(sinh(x)*cosh(x))dx = ∫udu = u^2/2 = (cosh(x)^2)/2 + C Verify d/dx((cosh(x)^2/)/2) = 2*cosh(x)*(sinh(x))/2 = sinh(x)*cosh(x) Formula 1. ∫(sinh(x)^n)*cos(x)dx = (cos(x)^(n+1))/(n+1) 2. ∫(cosh(x)^n)*sin(x)dx = (sin(x)^(n+1))/(n+1) Go to Begin Q14. ∫(sinh(2*x))dx = ? No anti-derivative is given Let u = 2*x and du = 2*dx Hence the integant is sinh(u)*(1/2)du ∫(sinh(2*x))dx = ∫(sinh(u)(1/2)du = cosh(u)/2 = cosh(2*x)/2 + C Verify d/dx(cosh(2*x)/2) = (2*sinh(2*x))/2 = sinh(2*x) Example : ∫(cosh(2*x))dx = ? No anti-derivative is given Let u = 2*x and du = 2*dx Hence the integant is cosh(u)*(1/2)du ∫(cosh(2*x))dx = ∫(cosh(u)(1/2)du = sinh(u)/2 = sinh(2*x)/2 + C Verify d/dx(sinh(2*x)/2) = (2*cosh(2*x))/2 = cosh(2*x) Formula 1. ∫(sinh(n*x))dx = cos(n*x)/(n+1) 2. ∫(cosh(n*x))dx = sin(n*x)/(n+1) Go to Begin

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