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inverse sinhx confusion

i've just used the substitution x=(1/3)sinh(u) in the integral √(1+9x^2) dx

its all gone fine (i think) until i try and put u=sinh^-1(3x) back into (1/12)sinhu

so i try to put ln(3x+√(9x^2 + 1)) in to (1/2)(e^x - e^-x)

but i cant get it to look like what the question says it should look like, which is (1/2)x√(1+9x^2).

what am i doing wrong?
Reply 1
Avatar for jpowell
jpowell
15
edm
i've just used the substitution x=(1/3)sinh(u) in the integral √(1+9x^2) dx

its all gone fine (i think) until i try and put u=sinh^-1(3x) back into (1/12)sinhu

so i try to put ln(3x+√(9x^2 + 1)) in to (1/2)(e^x - e^-x)

but i cant get it to look like what the question says it should look like, which is (1/2)x√(1+9x^2).

what am i doing wrong?


The actual answer is (1/2)x√(1+9x^2) + (1/6)Arcsinh[3x]

I = int(√(1+9x^2)dx)

Make the subs x = 1/3Sinh dx/du = 1/3Cosh

I = int(1/3Cosh^2du)

I = 1/6(u + CoshSinh)

I = 1/6(Arcsinh[3x] + 3x*Cosh[Arcsinh[3x]]) = (1/6)Arcsinh[3x] + (x/2)√(1+9x^2)
edm
i've just used the substitution x=(1/3)sinh(u) in the integral √(1+9x^2) dx

its all gone fine (i think) until i try and put u=sinh^-1(3x) back into (1/12)sinhu

so i try to put ln(3x+√(9x^2 + 1)) in to (1/2)(e^x - e^-x)

but i cant get it to look like what the question says it should look like, which is (1/2)x√(1+9x^2).

what am i doing wrong?

x=1/3sinhu => dx=1/3coshu
rt(1+9x^2)=rt(1+sinh^2u)=rt(cosh^2u)=coshu
integral becomes

int 1/3coshucoshu du
= 1/3int cosh^2u du
cos2u=2cosh^2u-1
hence integral is
1/6 int[cosh 2u+1]
=
1/6[1/2sinh2u+u]
1/6[coshusinhu+u]
1/6[rt(1+sinh^2u)sinhu+u]........using x=1/3sinhu => sinhu=3x
1/6[3xrt(1+9x^2)+sinh^(-1)(3x)]...using x=1/3sinh u=> 3x=sinhu=>sinh^(-1)3x=u
1/2xrt(1+9x^2)+1/6sinh^(-1)[3x]
Reply 3
Avatar for HighlyEvolved
HighlyEvolved
OP
1
evariste

1/6[1/2sinh2u+u]
1/6[coshusinhu+u]


thats the connection i wasnt making. thanks alot.

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