How do you integrate #int 1/sqrt(e^(2x)-2e^x+10)dx# using trigonometric substitution?

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Answer 1 · 2016-02-24T21:36:41

#int 1/sqrt(e^(2x)-2e^x+10)dx#

#int (e^xe^-x)/sqrt(e^(2x)-2e^x+10)dx#

#t = e^x#
#dt = e^x#

#int 1/(tsqrt(t^2-2t+10))dt#

#int 1/(tsqrt(t^2-2t+1 + 9))dt#

#int 1/(tsqrt((t-1)^2 + 9))dt#

#t-1 = 3tan(u)#
#dt = 3(tan^2(u)+1)du#

#(t-1)^2 = 9tan^2(u)#

#int 1/(tsqrt((t-1)^2 + 9))dx#

#int(3/cos^2(u))/((3tan(u)+1)(3/cos(u))) du#

#int(3/cos^2(u))/((9sin(u)/cos^2(u)+3/cos(u))) du#

#int1/((3sin(u)+cos(u))) du#

#w = tan(u/2)#

#3sin(u) = (6w)/(w^2+1)#

#cos(u) = (1-w^2)/(w^2+1)#

#du = (2dw)/(w^2+1) #

#int(2/(w^2+1))/(((6w)/(w^2+1)+(1-w^2)/(w^2+1))) dw#

#int2/(6w+1-w^2) dw#

#-int2/(w^2-6w+9-10) dw#

#-2int1/((w-3)^2-10) dw#

#sqrt(10)v = (w-3)#
#sqrt(10)dv = dw#

#-2sqrt(10)/10int1/(v^2-1) dv#

#-2sqrt(10)/10[arctanh(v)]+C#

#-2sqrt(10)/10[arctanh((tan(arctan((e^x-1)/3)/2)-3)/sqrt(10))]+C#