Question

integrate t sec^2 (2t) dt

Answer 1

integrate sec^2(2t) dt, integrate by parts, set dg=sec^2(2t)dt and g=(1/2)tan(2t), then you get (1/2) t tan(2t)-(1/2) integral (tan(2t) dt).
Do you manage the rest?

Answer 2

\[\int\limits_{}^{} t*\sec^{2}(2t)dt = t*\tan (2t) - \int\limits_{}^{}\tan (2t)dt \]

Answer 3

this is going to be:

= \[t*\tan (2t) + \ln |\cos (t)| + c\]

Answer 4

all of this was done by parts

Answer 5

all good except for typo in last line. it of course should be
\[\ln(\cos(2t))\]

Answer 6

in fact we are missing some constants too. should have
\[\frac{1}{2}t \tan(2t)+\frac{1}{4} \ln(\cos(2t))\] if i am not mistaken

Answer 7

@dumcow did i get this right? second constant i think is 1/4 because you need the 1/2 from first part when you do parts yes?

Answer 8

the constants always throw me off so i could easily be wrong

Answer 9

Satellite, I would agree with you over just about anyone on this site :) So I would probably agree your answer is right xP

Answer 10

And you're right if I'm doing this in my head correctly lol.