integrate tan^3(x)?

Answer 1

tan^3x=(1+sec^2x)tanx
=tanx+tanxsec^2x
integral of tanx=logsecx
for integral of tanxsec^2x,the substitution is tanx=t
sec^2x is the differential coefficient of tanx=dt
so the integral is t^2/2=>tan^2x/2
so integral tan^xdx=logsecx+tan^2x/2+C

Answer 2

Write the integrand as tan x(sec^2 x -1)
Now break the integral into 2 parts:
First int(tan x sec^2 x dx) is of the form int(u^2 du)
with u = tan x. So this part yields tan^2 x/2.
Finally, let's do int(tan x dx).
Write int(tan x dx )= int(sin x dx/ cos x). This is of the form
-int(du/u), with u = cos x.
So, our final result is
int(tan^3 x dx) = (tan^2 x)/2 + ln(cos x) + C

Answer 3

Alternative answer Log[Cos[x]] + Sec[x]^2/2

Answer 4

Here is a right answer u wanted

int of tan^3x
=int of tanx * tan^2x
=int of tanx(sec^2x-1)
=int of tanx.sec^2 - int of tanx
put tanx=1
by substitution method we differentiate & we get sec^x dx=dt.
=int of t dt - int of tanx dx.
=(t^2)/2-log(secx) + c
=(1/2)tan^2 - log(secx) + c
Thats the answer.

Answer 5

(1/2)*tan^2(x)+ln(|cosx|)+C
If you integrate by dx of course =)

Answer 6

log|cos x| + sec^2(x) / 2 + c