jaersyn
integrate (secx)^4 * sqrt (tanx) dx
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Outkast3r09
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I think you have to use Trig sub
Outkast3r09
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or you might be able to do this
Outkast3r09
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this is just a test idk if it will work
\[\int sec^2 x* sec^2 x*\sqrt{tanx} dx\]
Outkast3r09
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\[cos^2 x + sin^2 x= 1\]
\[1+tan^2 x=sec^2 x\]
jaersyn
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you're correct, i'm close, but not exactly the ansewr
Outkast3r09
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\[\int sec^2 x * (tan^2 x + 1) * \sqrt{tan x}dx\]
jaersyn
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not trig sub, u-sub
jaersyn
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yeah then let u = tanx; du = (secx)^2
Outkast3r09
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you could use trig subs
Outkast3r09
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trig sub will work with any trig integral... it's just a pain for something that can be done easier
anyways
\[\int sec^2 x *(tan^{3/2} x+ \sqrt{tan^{1/2}x})\]
Outkast3r09
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\[\int sec^2 x*tan^{3/2}x dx + \int sec^2 x*tan^{1/2}xdx\]
Outkast3r09
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\[\int sec^2 x* (tanx)^{3/2} dx + \int sec^2 x*(tanx)^{1/2}\]
Outkast3r09
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let \[u=tanx\]
\[du=sec^2 x\]
\[\int u^{3/2}du+\int u^{1/2}du\]
jaersyn
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i see where i went wrong
Outkast3r09
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ok =]
jaersyn
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ty
jaersyn
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wait
jaersyn
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∫sec2x∗(tan3/2x+tan1/2x−−−−−−√)
jaersyn
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how does tan^2 (x) * tan^(1/2) x = tan^(3/2) (x)?
Outkast3r09
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ehh it's supposed to be a 5/2
Outkast3r09
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so your answer wil lbe 7/2
random231
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wow