Question

∫ (sec θ - tan θ)² dθ

Answer 1

solve $∫ (sec θ - tan θ)² dθ$

$=∫(sec^2-2 secθtanθ+tan^2 θ)dθ$

$=∫(1+tan^2 θ-(2sec θtanθ+tan^2 θ)dθ$

$=∫(1-2secθtanθ+2tan^2 θ)dθ$

$=∫dθ-2∫(2secθtanθdθ)+2∫(tan^2 θdθ)$

$=θ-2∫secθtanθdθ+∫tan^2 θdθ$

for

$∫secθtanθdθ, let u=secθ , du=secθtanθdθ$

$؞ ∫secθtanθdθ=∫du=u =secθ …(ii)$

and for

$\int tan ^2 \theta d\theta$ we define $ǀn =∫tan^2 θdθ=\frac {tan^{-1} θ}{n-1}-ǀ_{n-2}$

as the reduction formula

Thus

$ǀ_2=∫tan^2 θdθ= \frac {tanθ}{1} -ǀ_0$

$=tanθ-θ…(iii)$

Replacing (ii) and (iii) in (i)gives,

$=∫(secθ-tanθ)^2 dθ$

$=θ-2 sec θ+2(tanθ-θ)+c$

$=2tanθ-2secθ-θ+c$