Integral Calculus> integration of trigonometric functions...

integrate tanxtan2xtan3x dx

Kuldeep Mukherjee , 14 Years ago

Grade 12

3 Answers

Sameer Kumar Srivastava

Last Activity: 14 Years ago

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

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

tell me are you satisfied with my solution.

vikas askiitian expert

Last Activity: 14 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

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

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