How do you integrate int 1/sqrt(e^(2x)-2e^x+10)dx1e2x2ex+10dx using trigonometric substitution?

1 Answer
Feb 24, 2016

int 1/sqrt(e^(2x)-2e^x+10)dx1e2x2ex+10dx

int (e^xe^-x)/sqrt(e^(2x)-2e^x+10)dxexexe2x2ex+10dx

t = e^xt=ex
dt = e^xdt=ex

int 1/(tsqrt(t^2-2t+10))dt1tt22t+10dt

int 1/(tsqrt(t^2-2t+1 + 9))dt1tt22t+1+9dt

int 1/(tsqrt((t-1)^2 + 9))dt1t(t1)2+9dt

t-1 = 3tan(u)t1=3tan(u)
dt = 3(tan^2(u)+1)dudt=3(tan2(u)+1)du

(t-1)^2 = 9tan^2(u)(t1)2=9tan2(u)

int 1/(tsqrt((t-1)^2 + 9))dx1t(t1)2+9dx

int(3/cos^2(u))/((3tan(u)+1)(3/cos(u))) du3cos2(u)(3tan(u)+1)(3cos(u))du

int(3/cos^2(u))/((9sin(u)/cos^2(u)+3/cos(u))) du3cos2(u)(9sin(u)cos2(u)+3cos(u))du

int1/((3sin(u)+cos(u))) du1(3sin(u)+cos(u))du

w = tan(u/2)w=tan(u2)

3sin(u) = (6w)/(w^2+1)3sin(u)=6ww2+1

cos(u) = (1-w^2)/(w^2+1)cos(u)=1w2w2+1

du = (2dw)/(w^2+1) du=2dww2+1

int(2/(w^2+1))/(((6w)/(w^2+1)+(1-w^2)/(w^2+1))) dw2w2+1(6ww2+1+1w2w2+1)dw

int2/(6w+1-w^2) dw26w+1w2dw

-int2/(w^2-6w+9-10) dw2w26w+910dw

-2int1/((w-3)^2-10) dw21(w3)210dw

sqrt(10)v = (w-3)10v=(w3)
sqrt(10)dv = dw10dv=dw

-2sqrt(10)/10int1/(v^2-1) dv210101v21dv

-2sqrt(10)/10[arctanh(v)]+C21010[arctanh(v)]+C

-2sqrt(10)/10[arctanh((tan(arctan((e^x-1)/3)/2)-3)/sqrt(10))]+C21010⎢ ⎢ ⎢ ⎢ ⎢ ⎢arctanh⎜ ⎜ ⎜ ⎜ ⎜ ⎜tan(arctan(ex13)2)310⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎥ ⎥ ⎥ ⎥ ⎥ ⎥+C