Integrating tan^2(u)(sec(u))du: Is There a Clean Way?

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The integral of tan^2(u)(sec(u))du can be approached by converting it to an expression involving sine and cosine, specifically as ∫(sin^2(u)/cos^3(u))du. Integration by parts is suggested for this approach, requiring careful breakdown. A natural substitution, x → sinh(y), simplifies the integral of (x^2)/sqrt(x^2+1)dx to ∫sinh^2(y)dy. By considering related integrals and applying hyperbolic trigonometric identities, the integral can be resolved more easily. Ultimately, the solution involves substituting back to the original variable.
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whats the integral of tan^2(u)(sec(u))du?

i was trying to integrate
(x^2)/sqrt(x^2+1)dx, and came into that. it turns out pretty messy though, is there a clean way to do it?
 
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Hi, I'm usually inclined to convert an integral like that to one with only sin's and cos's:

\int\frac{\sin(u)^2}{\cos(u)^3}du

Integration by parts will work on this if you break it up properly.
 
\int \frac{x^{2}}{\sqrt{x^{2}+1}} dx =...?
Make the natural substitution:x\rightarrow \sinh y

U'll be gettin' \int \sinh^{2}y dy (1)
Consider the "sister integral" \int \cosh^{2}y dy (2)

Consider the two expressions obtained by:(2)+(1);(2)-(1).The two new integrals will be trivials since u can use the 2 formulae from hyperbolic trigonometry:
\cosh^{2}y-\sinh^{2}y=1;\cosh^{2}y+\sinh^{2}y= \cosh{2y}

In the end u can extract this integral \int \sinh^{2}y dy easily and then in the final result u'll have to make the substitution back
y\rightarrow \arg\sinh x

Daniel.
 
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