Mathematics Dictionary Dr. K. G. Shih

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Integral of inverse trigonometric functions

Subjects

Symbol Defintion Example : Sqr(x) is square root of x Q01 | - ∫arcsin(x)dx = x*arcsin(x) + Sqr(1 - x^2) Q02 | - ∫arccos(x)dx =x*arccos(x) - Sqr(1 - x^2) Q03 | - ∫arctan(x)dx = x*arctan(x) - ln(1 + x^2) Q04 | - ∫arccsc(x)dx = x*arccsc(x) + ln(x + Sqr(x^2-1) Q05 | - ∫arcsec(x)dx = x*arcsec(x) - ln(x + Sqr(x^2-1) Q06 | - ∫arccot(x)dx = x*arccot(x) + ln(1 + x^2) 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. ∫arcsin(x)dx = x*arcsin(x) + Sqr(1 - x^2) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arcsin(x) and du = (1/Sqr(1 - x^2))dx Integral = u*v - ∫v*du Integral = x*arcsin(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 ∫arcsin(x)dx = x*arcsin(x) + Sqr(1 - x^2) Go to Begin Q02. ∫ arccos(x)dx = x*arccos(x) - Sqr(1 - x^2) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arccos(x) and du = (-1/Sqr(1 - x^2))dx Integral = u*v - ∫v*du Integral = x*arccos(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 ∫arccos(x)dx = x*arccos(x) - Sqr(1 - x^2) Go to Begin Q03. ∫ arctan(x)dx = x*arctan(x) - (ln(1 + x^2))/2 Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arctan(x) and du = (1/(1 + x^2))dx Integral = u*v - ∫v*du Integral = x*arctan(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) Hence ∫arctan(x)dx = x*arctan(x) - ln(1 + x^2)/2 Go to Begin Q04. ∫ arccsc(x)dx = x*arccsc(x) + ln(x + Sqr(x^2 -1) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arccsc(x) and du = (-1/(x*Sqr(x^2 - 1)))dx Integral = u*v - ∫v*du Integral = x*arccsc(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 ∫arccsc(x)dx = x*arccsc(x) + Sqr(x^2 - 1) Verify d/dx(x*arccsc(x) + Sqr(x^2 - 1) = arccsc(x) + x*(-1/(x*Sqr(x^2 - 1))) + (2*x)/(2*Sqr(x^2 - 1) = arccsc(x) Verify : the formula in title d/dx(x*arccsc(x) + ln(x + Sqr(x^2 - 1)) = arccsc(x) + x/(-x*Sqr(x^2 - 1)) + (1/(x + Sqr(x^2 - 1))*(d/dx(x + Sqr(x^2 - 1) = arccsc(x) - 1/Sqr(x^2 - 1) + (1/(x + Sqr(x^2 - 1)))*(1 + x/Sqr(x^2 - 1)) = arccsc(x) - 1/Sqr(x^2 - 1) + (1/(x + Sqr(x^2 - 1)))*(Sqr(x^2 - 1) + x)/Sqr(x^2 - 1)) = arccsc(x) - 1/Sqr(x^2 - 1) + (1/Sqr(x^2 - 1)) = arccsc(x) Go to Begin Q05. ∫ arcsec(x)dx = x*arcsec(x) - ln(x + Sqr(x^2 -1) Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arcsec(x) and du = (1/(x*Sqr(x^2 - 1)))dx Integral = u*v - ∫v*du Integral = x*arcsec(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 ∫arcsec(x)dx = x*arcsec(x) - Sqr(x^2 - 1) Verify d/dx(x*arcsec(x) - Sqr(x^2 - 1) = arcsec(x) + x*(1/(x*Sqr(x^2 - 1))) - (2*x)/(2*Sqr(x^2 - 1) = arcsec(x) Verify : the formula in title d/dx(x*arcsec(x) - ln(x + Sqr(x^2 - 1)) = arcsec(x) + x/(x*Sqr(x^2 - 1)) - (1/(x + Sqr(x^2 - 1))*(d/dx(x + Sqr(x^2 - 1) = arcsec(x) + 1/Sqr(x^2 - 1) - (1/(x + Sqr(x^2 - 1)))*(1 + x/Sqr(x^2 - 1)) = arcsec(x) + 1/Sqr(x^2 - 1) - (1/(x + Sqr(x^2 - 1)))*(Sqr(x^2 - 1) + x)/Sqr(x^2 - 1)) = arcsec(x) + 1/Sqr(x^2 - 1) - (1/Sqr(x^2 - 1)) = arcsec(x) Go to Begin Q06. ∫ arccot(x)dx = x*arccot(x) + (ln(1 + x^2))/2 Proof : Use integral by part Integral by part Let dv = dx and v = x Let u = arccot(x) and du = (-1/(1 + x^2))dx Integral = u*v - ∫v*du Integral = x*arccot(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) Hence ∫arctan(x)dx = x*arctan(x) + ln(1 + x^2)/2 Go to Begin Q07. ∫(1/Sqr(1 - x^2))dx = arcsin(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 = arcsin(x) Second method Let x = sin(U) then dx = cos(U)dU and U = arcsin(x) Hence (1/Sqr(1 - x^2)*dx = (1/Sqr(1 - sin(U)^2))*(cos(U))dU Since 1 - sin(U)^2 = cos(U)^2 Hence (1/Sqr(1 - x^2)*dx = dU Hence ∫(1/Sqr(1 - x^2))dx = ∫dU = U = arcsin(x) + C Go to Begin Q08. ∫(-1/Sqr(1-x^2))dx = arccos(x) Use anti-derivative : d/dx(arccos(x)) = -1/Sqr(1 - x^2) d(arccos(x)) = (-1/Sqr(1 - x^2))dx Hence integral of (-1/Sqr(1 - x^2))dx = arccos(x) Second method Let x = cos(U) then dx = -sin(U)dU and U = arccos(x) Hence (-1/Sqr(1 - x^2)*dx = (-1/Sqr(1 - cos(U)^2))*(-sin(U))dU Since 1 - cos(U)^2 = sin(U)^2 Hence (-1/Sqr(1 - x^2)*dx = dU Hence ∫(-1/Sqr(1 - x^2))dx = ∫dU = U = arccos(x) + C Go to Begin Q09. ∫(1/(1 + x^2))dx = arctan(x) Use derivative d/dx(arctan(x)) = 1/(1 + x^2) d(arctan(x)) = (1/(1 + x^2))dx Hence integral of (1/(1+x^2))dx = arctan(x) Second method Let x = tan(U) then dx = sec(U)^2)dU and U = arctan(x) Hence (1/(1 + x^2)*dx = (1/(1 + tan(U)^2))*(sec(U)^2)dU Since 1 + tan(U)^2 = sec(U)^2 Hence (1/(1 + x^2)*dx = dU Hence ∫(1/(1 + x^2))dx = ∫dU = U = arctan(x) + C Go to Begin Q10. ∫(csc(x)^2)dx = ? There is anti-derivative. Since d/dx(cot(x)) = -csc(x)^2 Hence d(cot(x)) = (-csc(2*x)^2)dx ∫(-csc(x)^2)dx = cot(x) + C Verify d/dx(cot(x)) = -csc(x)^2 Go to Begin Q11. ∫(sec(x)^2)dx = ? There is anti-derivative. Since d/dx(tan(x)) = sec(x)^2 Hence d(tan(x)) = (sec(2*x)^2)dx ∫(sec(x)^2)dx = tan(x) + C Verify d/dx(tan(x)) = sec(x)^2 Go to Begin Q12. ∫(-1/(1 + x^2))dx = arccot(x) Use derivative d/dx(arccot(x)) = -1/(1 + x^2) d(arccot(x)) = (-1/(1 + x^2))dx Hence integral of (-1/(1+x^2))dx = arccot(x) Second method Let x = cot(U) then dx = (-csc(U)^2)dU and U = arccot(x) Hence (-1/(1 + x^2)*dx = (1/(1 + cot(U)^2))*(-csc(U)^2)dU Since 1 + tan(U)^2 = csc(U)^2 Hence (-1/(1 + x^2)*dx = dU Hence ∫(-1/(1 + x^2))dx = ∫dU = U = arccot(x) + C Go to Begin Q13. New Go to Begin

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