Question

$\int \tan x.\tan 2x.\tan 3xdx$ is equal to

• A. $\frac{1}{3}\log |\sec 3x|-\frac{1}{2}\log |\sec 2x|+\log |\sec x|+C$
• B. $\frac{1}{3}\log |\sec 3x|-\frac{1}{2}\log |\sec 2x|-\log |\sec x|+C$
• C. $\frac{1}{3}\log |\sec 3x|+\frac{1}{2}\log |\sec 2x|+\log |\sec x|+C$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{3}\log |\sec 3x|-\frac{1}{2}\log |\sec 2x|-\log |\sec x|+C$

$tan\left(3x\right)=tan(x+2x)$

$tan\left(3x\right)=\frac{tanx+tan2x}{1-tanx.tan2x}$

tan(3x) - tanx tan(2x) tan(3x) = tanx + tan(2x)

$\therefore$ tan x.tan(2x) tan(3x) = tan(3x) - tan(x) - tan(2x) .....(i)

$\int tanx.tan\left(2x\right).tan\left(3x\right)dx$

$=\int tan\left(3x\right)-tan\left(x\right)-tan\left(2x\right).dx$ (fomi)

$=\frac{1}{3}log|sec\left(3x\right)|-log|secx|-\frac{1}{2}log|sec\left(2x\right)|+c$