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integrate tanx*tan2x*tan3x dx
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.
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.
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
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
Dear Student,Please find below the solution to your problem.Let’s get some weapons from our math battle armory :-)x+2x=3xtan(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∈RNow 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∈RThanks and Regards
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