Question

integral of trig functions

Answer 1

\[\tan^5\theta \sec^7\theta \] is the question
i split it to (tan^3tan^2)sec^2sec^5 and not sure what to do further

Answer 2

i tried let sec^2=1/cos^2

Answer 3

and tan^2=sin^2/cos^2

Answer 4

then i got\[\tan^3\sec^5(\frac{ 1 }{\cos^2 }.\frac{ \sin^2 }{\cos^2 })\] please ignore the lack of theta

Answer 5

am i on the right track?

Answer 6

split tan^5 x as tan x (tan^2x)^2 = tan x (sec^2 x -1)^2
then plug in u = sec x
du =...

Answer 7

see if you get
integral u^6 (u^2-1)^2 du
i'll be back after some time.

Answer 8

im confused with the split, you split tan and i got that but what happened to sec^7?

Answer 9

sec^7 x split as sec x sec^6x

Answer 10

\(\Huge \sec^6x (\sec^2x-1)^2 [\sec x \tan xdx]\)

put u= sec x