Question

Calculate the follow INTEGRAL:

Answer 1

\[\int\limits_{}^{}(tg^3(x)+tg(x))dx\]

Answer 2

Are you given anything about g(x)?
use integration by parts.

Answer 3

what in the world is tg?
is it tangent?

Answer 4

@eliassaab @thomaster @texaschic101 @undeadknight26 @phi @PeterPan @kropot72 @Luigi0210 @Lena772 @mathslover @NaomiBell1997

Answer 5

tg is tan "In my country"

Answer 6

\[=\int\limits_{}^{}tg^3(x)+\int\limits_{}^{}tg(x)=\int\limits_{}^{}tg^2(x)*tg(x)+\int\limits_{}^{}tg(x)\]
\[=\int\limits_{}^{}(\sec^2(x)-1)tg(x)+\int\limits_{}^{}tg(x)=\int\limits_{}^{}\sec^2(x)tg(x)\]
Now as long as (tg(x))'=sec^2(x) we substitute tg(x) with u:
\[\int\limits_{}^{}(u)du=\frac{ tg^2(x) }{ 2 }+C\]
THANKS VERY MUCH GUYS FOR YOUR SUPPORT (SARCASM)

Answer 7

If just somebody can find a "simple" way to do it ?

Answer 8

to be fair, it did take you quite a while to respond to my minor query.
oh well, as long as you already have it done :)

Answer 9

and I don't get it, how simple is simple?
Like you wouldn't need to use substitution?

Answer 10

It was a hard substitution to do

Answer 11

also, no need to split the integral into two.
just factor out one of the tg(x)
\[\large\int [\tan^3(x) + \tan(x)]dx = \int \tan(x) [\tan^2(x)+1]dx = \int \tan(x)\sec^2(x)\]
heck, even a substitution of u = sec(x) would work XD

Answer 12

@PeterPan You're right