Question

Integrate the following function with respect to x

$\int \: tan2x \: tan(x + a) \: tan(x - a) \: dx$
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Answer 1

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

Given integral is

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

To evaluate this type of integrals, which is given in three multiples of tan together such that one angle is equals to sum of other two angles, we solve as follow.

From given statement, we have

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

So,

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

We know,

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

So, using this identity, we get

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

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

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

So,

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

can be rewritten as

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

We know,

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

So, using this result, we get

$\rm \: = \: \dfrac{1}{2}log |sec2x| - log |sec(x + a)| - log |sec(x - a)| + c$

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ADDITIONAL INFORMATION

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