 # integrate tanx*tan2x*tan3x dx

12 years ago

using tan3x=(tanx+tan2x)/1+tanxtan2x

we have the given integration as ∫(tanx+tan2x-tan3x)

which is equal to ln|secx|+1/2ln|sec2x|-1/3ln|sce3x|

tell me are you satisfied with my solution.

12 years ago

I = tanxtan2xtan3xdx              ..............1

tan3x = tan(x+2x) = tanx + tan2x/1-tanxtan2x

tan3x(1-tanxtan2x) = tanx + tan2x

tanxtan2xtan3x =  tan3x - tanx - tan2x  .................2

from 1 & 2

I =( tan3x -tanx -tan2x )dx

I = logsec3x/3 - logsec2x/2 - logsecx + C

=log[ (sec3x)1/3/(sec2x)1/2(secx) ]  + C

approve if u like it

2 years ago
Dear Student,

Let’s get some weapons from our math battle armory :-)
x+2x=3x
tan(A+B)=tan(A)+tan(B)1−tan(A)tan(B)
∫tan(x)dx=ln|sec(x)|+k,k∈R
∫tan(ax+b)dx=1aln|sec(ax+b)|+k,k,a,b∈R
Now let’s enter the battlefield !
tan(3x)=tan(x+2x)=tan(x)+tan(2x)1−tan(x)tan(2x)
Therefore, tan(3x)−tan(x)tan(2x)tan(3x)=tan(x)+tan(2x)
which finally gets us something to work with ! (i.e.)
tan(3x)−tan(2x)−tan(x)=tan(x)tan(2x)tan(3x)
No
∫tan(x)tan(2x)tan(3x)dx=∫tan(3x)−tan(2x)−tan(x)dx=13ln|sec(3x)|−12ln|sec(2x)|−ln|sec(x)|+k,k∈R

Thanks and Regards