Question

integration of tan³x dx​

Answer 1

Solution:

Given Integral:

$\displaystyle \tt \longrightarrow I = \int {tan}^{3}x \: dx$

Can be written as:

$\displaystyle \tt \longrightarrow I = \int tan \: x \cdot{tan}^{2}x \: dx$

$\displaystyle \tt \longrightarrow I = \int tan \: x \: ({sec}^{2}x - 1) \: dx$

$\displaystyle \tt \longrightarrow I = \int tan \: x \: {sec}^{2}x \: dx- \int tan \: x\: dx$

Now, let us assume that:

$\tt \longrightarrow u =tan \: x$

$\tt \longrightarrow du=sec^{2}x \: dx$

So, the first integral changes to:

$\displaystyle \tt \longrightarrow I = \int u \: du- \int tan \: x\: dx$

$\displaystyle \tt \longrightarrow I = \dfrac{ {u}^{2} }{2} - \int tan \: x\: dx + C$

$\displaystyle \tt \longrightarrow I = \dfrac{ {u}^{2} }{2} - ln( |cos \: x| ) + C$

Substituting back u = tan(x), we get:

$\displaystyle \tt \longrightarrow I = \dfrac{ {tan}^{2}x }{2} - ln( |cos \: x| ) + C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

Answer:

tan² x / 2 + log | cos x | + c

Explanation:

Given integral

∫ tan³ x dx

= ∫ tan x × tan² x dx

Use trignometric identity tan² x = sec² x - 1

= ∫ tan x( sec² x - 1 ) dx

= ∫( tan xsec² x - tan x) dx

= ∫ tan xsec² x - ∫tan x dx

Let tan x = t

Differenciate

⇒ sec² x dx = dt

= ∫ t dt - ∫ tan x dx

= t²/2 + log | cos x | + c

= tan² x / 2 + log | cos x | + c