Question

integrate sec^3t

Answer 1

\[u=\sec t\]\[dv=\sec^2tdt\]integrate by parts

Answer 2

it's a classic integral

Answer 3

so it becomes sec t tan t - integration of tant sect tant

Answer 4

@TuringTest

Answer 5

got it bingo !

Answer 6

Sorry to interrupt, but I would like to know if this works too..
\[\int sec^3tdt\]\[=\int (tan^2t+1)sectdt\]\[=\int (tan^2tsect+sect)dt\]\[=\int (tan^2tsect)dt+\int (sect)dt\]\[=\int tantd(sect)+\int (sect)dt\]\[=sect \tan t-\int sectd(tant)+\int (sect)dt\]\[=sect \tan t-\int sec^3tdt+\int (sect)dt\]So\[2\int sec^3tdt = secttant+\int sect dt\]\[\int sec^3tdt = \frac{1}{2}(secttant+\int sect dt)=...\]
Seems I'm doing it wrong though.

Answer 7

this is exactly what I have done :)...thanks again...i already awarded the best response :( sorry ! thanks alot rolypoly for showing the detailed steps :) :) @RolyPoly

Answer 8

What have you got for the answer?
Btw, never mind :)

Answer 9

1/2 sect tant + ln (sec t + tant )....

Thanks again !!

Answer 10

1/2 [sect tant + ln (sec t + tant )] +C
Wow! It works!!~