Question

just verifying
\[\int tsec^{2}2tdt=\frac{t}{2}tan2t-\frac{1}{4}ln|sec2t| +C\]

Answer 1

is this the integral??\[ \int t \; \sec^2 (2t) dt \]

Answer 2

yes

Answer 3

i did integration by parts

Answer 4

Hmm .. let's try it by integration by parts.
\[ \frac 12 \sec^2 (2t) t^2 - \int \int \sec^2 (2t) dt dt \\
\text{ Ignoring frst term}\\
\frac 12\int \tan (2t) dt = \frac 14 \ln |\sec (2t)|\]
I guess it should be
\[ \frac 12 \sec^2 (2t) t^2 - \frac 14 \ln |\sec (2t)|\]

Answer 5

let's try asking wolf

Answer 6

what went wrong
http://www.wolframalpha.com/input/?i=int+t+sec^2+%282t%29

Answer 7

I disagree with wolf

Answer 8

oh ... i got the wrong formula for integration by parts.
the first term should be
\[ t \int \sec^2 (2t) = \frac 12t \tan (2t) \]

Answer 9

why??

Answer 10

it seem you are right ... it's just manipulation of log
- log(sec x) = log(cos x)

Answer 11

lets see
u=t dv=sec^(2) 2t dt
du=dt v=(1/2)tan2t
\[uv-\int vdu\]
\[\int tsec^{2}2tdt=\frac{t}{2}tan2t-\frac{1}{4}ln|sec2t| +C\]

Answer 12

\[ - \frac 14 \ln |\sec (2t)| = \frac 14 \ln \left |\frac 1{\sec (2t)} \right | = \frac 14 \ln |\cos (2t)|\]

Answer 13

oh....i see

Answer 14

both answers are equivalent.

Answer 15

makes sense.. thanks again experimentX!

Answer 16

you are welcome :)