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Mathematics 22 Online
OpenStudy (lgbasallote):

This is not a riddle! well kinda riddlish >:)) how to integrate \(\LARGE \int{}{} \frac{1}{\cos x} dx\) step-by-step

OpenStudy (blockcolder):

I was gonna say Weierstrass substitution but meh, too long.

OpenStudy (anonymous):

idk :'( I can use integration by parts but @Mimi_x3 has a better way to do it.

OpenStudy (lgbasallote):

@Mimi_x3 is a calculus expert huh

OpenStudy (blockcolder):

Turn 1/cos(x) to sec(x) and multiply both numerator and denominator by sec(x)+tan(x).

OpenStudy (anonymous):

OpenStudy (anonymous):

Yeah, that's what Mimi does. The blockcoder way.

OpenStudy (lgbasallote):

then \(\LARGE \frac{\sec^{2} x + secxtanx}{secxtanx}\)

OpenStudy (lgbasallote):

what does that mean o.O

OpenStudy (lgbasallote):

let u = secx tanx ??

OpenStudy (blockcolder):

There's a missing + in the denominator. =))

OpenStudy (callisto):

\[\int secx \frac {secx+tanx}{secx+tanx} dx\]\[=\int \frac {sec^x+secxtanx}{secx+tanx} dx\]\[=\int \frac {1}{secx+tanx} d(secx+tanx)\] =ln |secx + tanx| +C

OpenStudy (lgbasallote):

where'd that second to the last line come from again?

OpenStudy (callisto):

\[\frac{d}{dx} (secx + tanx) = sec^2x + secxtanx\]

OpenStudy (lgbasallote):

oh i see

OpenStudy (lgbasallote):

so it was let u = secx + tanx right...

OpenStudy (callisto):

Perhaps... It should be... I don't know... I just did it in this way. Sorry!!!

OpenStudy (lgbasallote):

i like u substitution :P hmm..so this doesnt have the trig substitution yet...gotta find one

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