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\[\int\limits \frac{ secxtanx }{ 4+\sec^2x }\]
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im pretty much lost with the trig functions
i know i have to use inverse functions but secxtanx, is the inverse just secx?
a simple substitution of secx = t will help your case.
could you show me?
secx = t secx tanx dx = dt rest is easier
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so is it u/4+sec^2x ?
nop o.O it becomes dt/(4 + t^2)
\[\frac{ secxtanx }{ 4 + (secxtanx)^2 }= 1/4 +\sec x tanx\]
is that my answer or am i still way off?
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