Mathematics Dictionary Dr. K. G. Shih

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Integral : Inverse hyperbolic functions

Subjects

Symbol Defintion Example : Sqr(x) is square root of x Q01 | - ∫arcsinh(x)dx = x*arcsinh(x) - Sqr(1 + x^2) Q02 | - ∫arccosh(x)dx =x*arccosh(x) - Sqr(x^2 - 1) Q03 | - ∫arctanh(x)dx = x*arctanh(x) - ln(1 - x^2) Q04 | - ∫arccsch(x)dx = x*arccsch(x) - ln(x + Sqr(x^2 + 1) Q05 | - ∫arcsech(x)dx = x*arcsech(x) - ln(x + Sqr(1 - x^2) Q06 | - ∫arccot(x)dx = x*arccot(x) + ln(x^2 - 1) Q07 | - ∫(1/Sqr(1 - x^2))dx = arcsin(x) Q08 | - ∫(-1/Sqr(1 - x^2))dx = arccos(x) Q09 | - ∫(1/(1 + x^2))dx =arctan(x) Q10 | - ∫(-1/(x*Sqr(x^2 - 1)))dx = arccsc(x) Q11 | - ∫(1/(x*Sqr(x^2 - 1)))dx = arcsec(x) Q12 | - ∫(-/(1 + x^2))dx = arccot(x) Q13 | - ∫(sin(x)*cos(x))dx = ? Q14 | - ∫(sin(2*x))dx = ?

Answers

Q01. ∫arcsinh(x)dx = x*arcsinh(x)- Sqr(1 + x^2) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arcsinh(x) and du = (1/Sqr(1 + x^2))dx Integral = u*v - ∫v*du Integral = x*arcsinh(x) - ∫(x/Sqr(1+x^2))dx Find Integral = ∫(x/Sqr(1+x^2))dx Let w = 1 + x^2 and dw = 2*xdx Hence integrant = (1/(2*Sqr(w))dw = (1/2)*(1/(-1/2+1))Sqr(w) = Sqr(1 + x^2) Hence ∫arcsinh(x)dx = x*arcsinh(x) - Sqr(1 + x^2) Go to Begin Q02. ∫ arccosh(x)dx = x*arccosh(x) - Sqr(x^2 - 1) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arccosh(x) and du = (-1/Sqr(x^2 - 1))dx Integral = u*v - ∫v*du Integral = x*arccosh(x) - ∫(-x/Sqr(x^2 - 1))dx Find Integral = ∫(-x/Sqr(x^2 - 1))dx Let w = x^2 - 1 and dw = 2*xdx Hence integrant = (-1/(2*Sqr(w))dw = (-1/2)*(1/(-1/2+1)Sqr(w) = Sqr(x^2 - 1) Hence ∫arccosh(x)dx = x*arccosh(x) - Sqr(x^2 - 1) Go to Begin Q03. ∫ arctanh(x)dx = x*arctanh(x) - (ln(1 - x^2))/2 Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arctanh(x) and du = (1/(1 - x^2))dx Integral = u*v - ∫v*du Integral = x*arctanh(x) - ∫(x/(1 - x^2))dx Find Integral = ∫(x/(1 - x^2))dx Let w = x^2 - 1 and dw = 2*xdx Hence integrant = (1/2)*(1/(w))dw = (1/2)*(ln(w) = ln(1 - x^2)/2 Hence ∫arctanh(x)dx = x*arctanh(x) - ln(1 - x^2)/2 Go to Begin Q04. ∫ arccsch(x)dx = x*arccsch(x) - ln(x + Sqr(x^2 + 1) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arccsch(x) and du = (-1/(x*Sqr(x^2 + 1)))dx Integral = u*v - ∫v*du Integral = x*arccsch(x) - ∫(-1/2)/Sqr(x^2 + 1))dx Find Integral = ∫(1/2)/Sqr(x^2 - 1))dx Let w = x^2 + 1 and dw = 2*xdx Hence integrant = (1/(2*Sqr(w))dw = (1/2)*(1/(-1/2+1))*Sqr(w) = Sqr(x^2 + 1) Hence ∫arccsch(x)dx = x*arccsch(x) - Sqr(x^2 + 1) Also ∫arctanh(x)dx = x*arccsch(x) + arcsinh(x) Verify d/dx(x*arccsch(x) + Sqr(x^2 - 1) = arccsch(x) + x*(-1/(x*Sqr(x^2 - 1))) + (2*x)/(2*Sqr(x^2 - 1) = arccsch(x) Verify : the formula in title d/dx(x*arccsc(x) + ln(x + Sqr(x^2 + 1)) = arccsch(x) + x/(-x*Sqr(x^2 - 1)) + (1/(x + Sqr(x^2 - 1))*(d/dx(x + Sqr(x^2 - 1) = arccsch(x) - 1/Sqr(x^2 - 1) + (1/(x + Sqr(x^2 - 1)))*(1 + x/Sqr(x^2 - 1)) = arccsch(x) - 1/Sqr(x^2 - 1) + (1/(x + Sqr(x^2 - 1)))*(Sqr(x^2 - 1) + x)/Sqr(x^2 - 1)) = arccsch(x) - 1/Sqr(x^2 - 1) + (1/Sqr(x^2 - 1)) = arccsch(x) Go to Begin Q05. ∫ arcsech(x)dx = x*arcsech(x) - ln(x + Sqr(1 - x^2) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arcsech(x) and du = (-1/(x*Sqr(1 - x^2)))dx Integral = u*v - ∫v*du Integral = x*arcsech(x) - ∫(1/2)/Sqr(1 - x^2))dx Find Integral = ∫(1/2)/Sqr(1 - x^2))dx Let w = 1 - x^2 and dw = -2*xdx Hence integrant = (1/(2*Sqr(w))dw = (1/2)*(1/(-1/2+1))*Sqr(w) = Sqr(1 - x^2) Hence ∫arcsec(x)dx = x*arcsec(x) - Sqr(1 - x^2) Also ∫arctanh(x)dx = x*arcsech(x) + arcsin(x) Verify d/dx(x*arcsech(x) - Sqr(1 - x^2) = arcsech(x) + x*(-1/(x*Sqr(x^2 - 1))) - (2*x)/(2*Sqr(1 - x^2) = arcsech(x) Verify : the formula in title d/dx(x*arcsech(x) - ln(x + Sqr(1 - x^2)) = arcsech(x) + x/(x*Sqr(1 - x^2)) - (1/(x + Sqr(1 - x^2))*(d/dx(x + Sqr(1 - x^2) = arcsech(x) + 1/Sqr(1 - x^2) - (1/(x + Sqr(1 - x^2)))*(1 + x/Sqr(x^2 - 1)) = arcsech(x) + 1/Sqr(1 - x^2) - (1/(x + Sqr(1 - x^2)))*(Sqr(1 - x^2) + x)/Sqr(1 - x^2)) = arcsech(x) + 1/Sqr(1 - x^2) - (1/Sqr(1 - x^2)) = arcsech(x) Go to Begin Q06. ∫ arccoth(x)dx = x*arccoth(x) + (ln(1 - x^2))/2 Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arccoth(x) and du = (1/(1 - x^2))dx Integral = u*v - ∫v*du Integral = x*arccoth(x) - ∫(x/(1 - x^2))dx Find Integral = ∫(x/(1 - x^2))dx Let w = 1 - x^2 and dw = -2*xdx Hence integrant = (-1/2)*(1/(w))dw = (-1/2)*(ln(w) = -ln(1 - x^2)/2 Hence ∫arccoth(x)dx = x*arccoth(x) + ln(1 - x^2)/2 Go to Begin Q07. ∫(1/Sqr(1 + x^2))dx = arcsinh(x) Use anti-derivative : d/dx(arcsin(x)) = 1/Sqr(1 + x^2) d(arcsin(x)) = (1/Sqr(1 + x^2))dx Hence integral of (1/Sqr(1 + x^2))dx = arcsinh(x) Second method Let x = sinh(U) then dx = cosh(U)dU and U = arcsinh(x) Hence (1/Sqr(1 + x^2)*dx = (1/Sqr(1 + sinh(U)^2))*(cosh(U))dU Since cosh(U)^2 - sinh(U)^2 = 1 Hence (1/Sqr(1 + x^2)*dx = dU Hence ∫(1/Sqr(1 + x^2))dx = ∫dU = U = arcsinh(x) + C Go to Begin Q08. ∫(1/Sqr(x^2 - 1))dx = arccosh(x) Use anti-derivative : d/dx(arccosh(x)) = 1/Sqr(x^2 - 1) d(arccosh(x)) = (1/Sqr(x^2 - 1))dx Hence integral of (1/Sqr(x^2 - 1))dx = arccosh(x) Second method Let x = cosh(U) then dx = sinh(U)dU and U = arccosh(x) Hence (1/Sqr(x^2 - 1)*dx = (1/Sqr(cos(U)^2 - 1))*(-sinh(U))dU Since cosh(U)^2 - 1 = sinh(U)^2 Hence (1/Sqr(x^2 - 1)*dx = dU Hence ∫(1/Sqr(x^2 - 1))dx = ∫dU = U = arccosh(x) + C Go to Begin Q09. ∫(1/(1 - x^2))dx = arctanh(x) Use derivative d/dx(arctanh(x)) = 1/(1 - x^2) d(arctanh(x)) = (1/(1 - x^2))dx Hence integral of (1/(1 - x^2))dx = arctanh(x) Second method Let x = tanh(U) then dx = sech(U)^2)dU and U = arctanh(x) Hence (1/(1 - x^2)*dx = (1/(1 - tanh(U)^2))*(sech(U)^2)dU Since 1 - tanh(U)^2 = sech(U)^2 Hence (1/(1 + x^2)*dx = dU Hence ∫(1/(1 - x^2))dx = ∫dU = U = arctanh(x) + C 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 ∫(-csc(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. ∫(1/(1 - x^2))dx = arccoth(x) Use derivative d/dx(arccoth(x)) = 1/(1 - x^2) d(arccoth(x)) = (1/(1 - x^2))dx Hence integral of (1/(1 - x^2))dx = arccoth(x) Second method Let x = coth(U) then dx = (-csch(U)^2)dU and U = arccoth(x) Hence (-1/(1 - x^2)*dx = (1/(1 - coth(U)^2))*(-csch(U)^2)dU Since 1 - coth(U)^2 = csch(U)^2 Hence (-1/(1 - x^2)*dx = dU Hence ∫(-1/(1 - x^2))dx = ∫dU = U = arccoth(x) + C Go to Begin Q13. New Go to Begin

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