Question

\[\int tan^2(t)dt\]

Answer 1

the owlfred is white .... you spose to be green or they change that?

Answer 2

\[sin^2(t) + cos^2(t) = 1\]
\[\implies tan^2(t) + 1 = sec^2(t)\]\[\implies tan^2(t) = \frac{1}{cos^2(t)} - 1\]

Answer 3

I think we need that one if I remember right.

Answer 4

maybe, i had considered using a useful form of 0 tho; like +1-1

Answer 5

gets the same results

Answer 6

\[\int tan^2(t)+1-1dt\]
\[\int sec^2(t)-1dt\]

Answer 7

And we need the one I can never recall:
\[cos^2(t) = \frac{cos(2t) + 1}{2}\]

Answer 8

Oh. Actually we don't need that first one.

Answer 9

It's this last one that's the important bit

Answer 10

the reduction in powers is a good thing to remember fer sure

Answer 11

better you than me!

Answer 12

if its done right, i think it gets to
\[\int tan^2(t)dt=tan(t)-t+C\]

Answer 13

i think without the square

Answer 14

oh right, what you wrote

Answer 15

reminiscent of
\[\int log(x)dx=x\log(x)-x\]

Answer 16

yep

Answer 17

hmm maybe parts works instead of reduction formula?

Answer 18

nah, just +1-1 and stab away

Answer 19

\[\int \sin^2(x)sec^2(x)dx\]?

Answer 20

that would be fun to throw at the kids lol

Answer 21

got a nice dv there right?

Answer 22

i am guessing but might work

Answer 23

there is no nice dv in that one

Answer 24

\[\sec^2(x)=dv\]
\[v=\tan(x)\]

Answer 25

how does that make an easier intergral from the first one tho?

Answer 26

ignore me i am just guessing.

Answer 27

the simplest way i saw, is to add 0; and solve

\[\int tan^2(t)dt => \int tan^2+1-1dt\]
\[\int (tan^2(t)+1)-(1)dt =>\int sec^2(t) -1 = tan(t)-t+C\]

Answer 28

yeah that looks nice and easy. easier still is to go right to
\[\tan^2(x)=\sec^2(x)-1\] and take anti derivatives

Answer 29

lol ... yeah, but that was way to easy :)