Question

Final one for today - Promise xD
@solomonzelman

Answer 1

Yes, perfect, you got it!

Answer 2

AWESOME. :)

Answer 3

Finally done with my homework. Whew! Now I can go eat dinner.
Thanks a million!

Answer 4

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \sec^{2m}z~dz;\quad m\in\mathbb{N},~m\le 2}\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} (1+\tan^{2(m-1)}z)\sec^2z~dz}\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} 1+u^{2(m-1)}~du=u+\frac{u^{2m-1}}{2m-1}+C=\tan z+\frac{\tan ^{2m-1}z}{2m-1}+C}\)

Answer 5

Generalization.

Answer 6

Sometimes, perhaps it is better to generalize. It's precise, and the math spirit rises.

Answer 7

why must m <= 2?

Answer 8

Well,
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \sec^{2}z~dz=\tan x+C}\)

Answer 9

oh tan(z) Not x, but the point is the same...

Answer 10

oh I meant m>=2.
Sorry

Answer 11

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \sec^{2m}z~dz=\tan z+\frac{\tan ^{2m-1}z}{2m-1}+C\quad {\small \rm when,}\quad \left\{m|m\in \mathbb{N},m\ge 2\right\}}\)

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \sec^{2 m}z~dz=\tan z+C\quad {\small \rm when}\quad m=1}\)

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \sec^{2 m}z~dz=z+C\quad {\small \rm when}\quad m=0}\)

Answer 12

Ah, I see now.

Answer 13

and for negative integer m, they are cosines to positive powers...

Answer 14

Don't want to overwhelm with generalizations though; good luck with integration!@

Answer 15

Thanks! Have a good night (or whatever it is wher you live)

Answer 16

I live in Atlanta GA, yes night:)
.... \(u_2\)

Answer 17

lol, it's late where you are. 3 hrs later than where I am