Question

what is the integral of tan^5 (2x)

Answer 1

hi sarah :)
did you try any substitution?

Answer 2

hint :
start by separating tan^2 (2x) and writing it as (1+ sec^2 2x)
:)

Answer 3

yes i tried and i got u4\4 - u2\2 + ln(cos2x) + C is that rigt?

Answer 4

thats somewhat correct...
your u is sec 2x , right ?
so re substitute u = sec 2x in 1st two terms

and for the 3rd term i am getting -4log (cos 2x)
can you check your working again ? or show it here so that i can help you spot the error :)

Answer 5

Stan^32xtan^22xdx
Stan^32x(sec^22x-1)dx
Stan^32xsec^22x-Stan^32Xdx
Su^3du - Stan2xtan^22xdx
Su^3du-Stan2x(sec^22x-1)dx
Su^3du - Stan2xsecx^22xdx - Stan2xdx
Su^3du - Sudu -Stan2xdx
u4/4 -u2/2 + ln|cos2x|+C
tan^42x/4 - tan^22x/2 + ln|cos2x| + c

Answer 6

give me some time to check all steps,
also,
if you plug in u = tan 2x
du is actually
\(\large 2\)sec^2(2x)

Answer 7

also
-tan 2x times -1 is +tan 2x
you've taken it as - tan 2x

Answer 8

these are the only 2 errors in your working :)

Answer 9

after correcting those,
see if you get your finals answer as
\((1/8)\tan^42x -(1/4)\tan^2 2x -(1/2)\ln (\cos 2x)+c\)
:)

Answer 10

oh and bdw,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 11

THANKKKKKK YOUUUUU :)))))))

Answer 12

you're welcome ^_^