- #26

vela

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Put this

\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

Finally, undo the substitution

together with this[tex] \int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2} [/tex]

To getSo, now

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

\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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