Question

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

Answer 1

I think you have to use Trig sub

Answer 2

or you might be able to do this

Answer 3

this is just a test idk if it will work

\[\int sec^2 x* sec^2 x*\sqrt{tanx} dx\]

Answer 4

\[cos^2 x + sin^2 x= 1\]
\[1+tan^2 x=sec^2 x\]

Answer 5

you're correct, i'm close, but not exactly the ansewr

Answer 6

\[\int sec^2 x * (tan^2 x + 1) * \sqrt{tan x}dx\]

Answer 7

not trig sub, u-sub

Answer 8

yeah then let u = tanx; du = (secx)^2

Answer 9

you could use trig subs

Answer 10

trig sub will work with any trig integral... it's just a pain for something that can be done easier

anyways

\[\int sec^2 x *(tan^{3/2} x+ \sqrt{tan^{1/2}x})\]

Answer 11

\[\int sec^2 x*tan^{3/2}x dx + \int sec^2 x*tan^{1/2}xdx\]

Answer 12

\[\int sec^2 x* (tanx)^{3/2} dx + \int sec^2 x*(tanx)^{1/2}\]

Answer 13

let \[u=tanx\]
\[du=sec^2 x\]

\[\int u^{3/2}du+\int u^{1/2}du\]

Answer 14

i see where i went wrong

Answer 15

ok =]

Answer 16

ty

Answer 17

wait

Answer 18

∫sec2x∗(tan3/2x+tan1/2x−−−−−−√)

Answer 19

how does tan^2 (x) * tan^(1/2) x = tan^(3/2) (x)?

Answer 20

ehh it's supposed to be a 5/2

Answer 21

so your answer wil lbe 7/2

Answer 22

wow