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Ray Vickson
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QuarkCharmer said:Related to this question:
[tex]\int sec^{3}(x)tan(x) dx[/tex]
[itex]u=sec(x)[/itex], and so, [itex]\frac{du}{dx}=sec(x)tan(x)[/itex], and [tex]dx=\frac{du}{sec(x)tan(x)}[/tex]
[tex]\int u^{3}tan(x) dx[/tex]
[tex]\int \frac{u^{3}tan(x)}{sec(x)tan(x)} du[/tex]
[tex]\int \frac{u^{3}}{sec(x)} du[/tex]
I have no idea what to do with this one? Can I put my u=sec back in and try again now?
Edit: Na, that just gets me back to where I started.
Oh wait, because u is equal to sec, can I just call that [itex]\frac{u^{3}}{u}[/itex] ?
You could note that [itex] f(x) = \sec^{3}(x) \tan(x)[/itex] has the form [itex] f(x) = \sin(x)/ \cos^{4}(x) [/itex], so substituting [itex] u = \cos(x) [/itex] gives [itex] f(x) dx = - du/u^4 [/itex].
RGV