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Integral of sech^3 (x)

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  • #26
vela
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Put this
[tex] \int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2} [/tex]
together with this
So, now
[tex]\int \frac{du}{(1+u^2)^2} = \frac{\arctan u}{2} + \frac{\sin(\arctan u)\cos(\arctan u)}{2}\; .......equation 1[/tex]
To get
\begin{align*}
\int \text{sech}^3 x\,dx &= \int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2} \\
&= \frac{\arctan u}{2} + \frac{\sin(\arctan u)\cos(\arctan u)}{2}\end{align*}Forget the rest of the stuff.

Draw a right triangle with the leg opposite θ having length u and the adjacent leg having length 1. Then you'd have tan θ = u/1 = u. What's the length of the hypotenuse of this triangle in terms of u? Once you have that, write down algebraic expressions for sin(arctan u) = sin θ and cos(arctan u) = cos θ in terms of u.

Finally, undo the substitution u = sinh x to obtain the final answer. Don't forget to tack on the arbitrary constant of integration at the end.
 
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  • #27
vela
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In the other approach, you had
[tex]\int\frac{1}{1+u^2}\,du = \frac{u}{1+u^2}+2\int\frac{1}{1+u^2}\,du - 2\int\frac{1}{(1+u^2)^2}\,du[/tex]
If you let [itex]I = \int\frac{1}{(1+u^2)^2}\,du[/itex], you can say
[tex]\int\frac{1}{1+u^2}\,du = \frac{u}{1+u^2}+2\int\frac{1}{1+u^2}\,du - 2I[/tex]Now you solve this for I and evaluate the remaining integrals, you'll find I equals the same expression in terms of u that you'll get after simplifying sin(arctan u) and cos(arctan u).
 
  • #28
DryRun
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Draw a right triangle with the leg opposite θ having length u and the adjacent leg having length 1. Then you'd have tan θ = u/1 = u. What's the length of the hypotenuse of this triangle in terms of u? Once you have that, write down algebraic expressions for sin(arctan u) = sin θ and cos(arctan u) = cos θ in terms of u.

Finally, undo the substitution u = sinh x to obtain the final answer. Don't forget to tack on the arbitrary constant of integration at the end.
The hypotenuse of the triangle is: [tex]\sqrt{1 + u^2}[/tex]

[tex]\int \frac{du}{(1+u^2)^2} = \frac{\arctan u}{2} + \frac{\sin(\arctan u)\cos(\arctan u)}{2}[/tex]

[tex]\frac{\arctan u}{2} + \frac{\sin(\arctan u)\cos(\arctan u)}{2} = \frac{\theta}{2} + \frac{\sin \theta\cos\theta}{2} = \frac{\arctan u}{2} + \frac{u}{2(1 + u^2)}[/tex]

[tex]\frac{\arctan u}{2} + \frac{u}{2(1 + u^2)} = \frac{\arctan (sinh x)}{2} + \frac{\sinh x}{2(1 + \sinh^2 x)} = \frac{\arctan (\sinh x)}{2} + \frac{(sech x)(\tanh x)}{2} + C[/tex]

OK, this is very close to the final answer, but this part does not match the answer in my notes:
[tex]\frac{\arctan (\sinh x)}{2}[/tex]
It should be:
[tex]\arctan (\tanh \frac{1}{2}x)[/tex]
 
  • #29
dextercioby
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They should be equal up to a numerical constant.
 
  • #30
DryRun
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Is there a way to convert one into the other? I ask because i should give the answer exactly.
 
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  • #31
dextercioby
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If 2 functions have the same derivative, then they must be equal up to a constant...
 

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