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integrate (secx)^4 * sqrt (tanx) dx

Mathematics
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I think you have to use Trig sub
or you might be able to do this
this is just a test idk if it will work \[\int sec^2 x* sec^2 x*\sqrt{tanx} dx\]

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Other answers:

\[cos^2 x + sin^2 x= 1\] \[1+tan^2 x=sec^2 x\]
you're correct, i'm close, but not exactly the ansewr
\[\int sec^2 x * (tan^2 x + 1) * \sqrt{tan x}dx\]
not trig sub, u-sub
yeah then let u = tanx; du = (secx)^2
you could use trig subs
trig sub will work with any trig integral... it's just a pain for something that can be done easier anyways \[\int sec^2 x *(tan^{3/2} x+ \sqrt{tan^{1/2}x})\]
\[\int sec^2 x*tan^{3/2}x dx + \int sec^2 x*tan^{1/2}xdx\]
\[\int sec^2 x* (tanx)^{3/2} dx + \int sec^2 x*(tanx)^{1/2}\]
let \[u=tanx\] \[du=sec^2 x\] \[\int u^{3/2}du+\int u^{1/2}du\]
i see where i went wrong
ok =]
ty
wait
∫sec2x∗(tan3/2x+tan1/2x−−−−−−√)
how does tan^2 (x) * tan^(1/2) x = tan^(3/2) (x)?
ehh it's supposed to be a 5/2
so your answer wil lbe 7/2
wow

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