Question

Integrate with respect to x

tan2x tan (x-a) tan(x+a)

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm tan2x \: tan(x - a) \: tan(x + a) \: dx$

Now, to evaluate this integral,

We know,

$\rm \: 2x = x + a + x - a$

$\rm\implies \:tan2x = tan[(x + a) + (x - a)]$

$\rm \: tan2x = \dfrac{tan(x + a) + tan(x - a)}{1 - tan(x + a) \: tan(x - a)}$

$\rm \: tan2x - tan2xtan(x + a)tan(x - a) = tan(x + a) + tan(x - a)$

So,

$\rm \: tan2xtan(x + a)tan(x - a) = tan2x - tan(x + a) - tan(x - a)$

Hence, given integral

$\rm \: \displaystyle\int\rm tan2x \: tan(x - a) \: tan(x + a) \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm[ tan2x - \: tan(x - a) - \: tan(x + a) ]\: dx$

$\rm \: = \: \dfrac{log |sec2x| }{2} - log |sec(x - a)| - log |sec(x + a)| + c$

Hence,

$\color{green}\rm\implies \:\displaystyle\int\rm tan2x \: tan(x - a) \: tan(x + a) \: dx \\ \\ \color{green}\rm \: = \: \dfrac{log |sec2x| }{2} - log |sec(x - a)| - log |sec(x + a)| + c$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }}$

$\boxed{ \rm{ \:\displaystyle\int\rm tanx \: dx \: = \: log |secx| \: + \: c \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

Let I=∫tan(x−a)tan(x+a)tan2xdx

As 2x=(x+a)+(x−a)⇒tan2x=

1−tan(x+a)tan(x−a)

tan(x+a)+tan(x−a)

⇒tan2x−tan(x+a)−tan(x−a)=tan2xtan(x+a)tan(x−a)

Therefore

I=∫[tan2x−tan(x+a)−tan(x−a)]dx

=

2

1

logsec2x−logsec(x+a)−logsec(x−a)

=log{

sec2x

cos(x+a)cos(x−a)}.

hope it helps

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