Question

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

Answer 1

`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`