How do you integrate $({sec (x)}^{2} - 1) (\frac{sin (x)}{cos (x)})$ $(sec(x)^2-1)(sin(x)/cos(x))$ ?

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Answer 1 · 2017-01-13T06:19:51

$\int ({sec}^{2} x - 1) (\frac{sin x}{cos x}) d x = \frac{{tan}^{2} x}{2} + ln cos x + c$ $int(sec^2x-1)(sinx/cosx)dx=tan^2x/2+lncosx+c$

$\int ({sec}^{2} x - 1) (\frac{sin x}{cos x}) d x$ $int(sec^2x-1)(sinx/cosx)dx$

$= \int ({sec}^{2} x \times \frac{sin x}{cos x}) d x - \int \frac{sin x}{cos x} d x$ $=int(sec^2x xxsinx/cosx)dx-intsinx/cosxdx$

$= \int {sec}^{2} x tan x d x - \int \frac{sin x}{cos x} d x$ $=intsec^2xtanxdx-intsinx/cosxdx$

Let us first solve $\int {sec}^{2} x tan x d x$ $intsec^2xtanxdx$

let $u = tan x$ $u=tanx$ , then $d u = {sec}^{2} x d x$ $du=sec^2xdx$

Hence $\int {sec}^{2} x tan x d x = \int u d u$ $intsec^2xtanxdx=intudu$

$= \frac{u^{2}}{2} = \frac{{tan}^{2} x}{2}$ $=u^2/2=tan^2x/2$

For $\int \frac{sin x}{cos x} d x$ $intsinx/cosxdx$ , let $v = cos x$ $v=cosx$ than $d v = - sin x d x$ $dv=-sinxdx$ and

$\int \frac{sin x}{cos x} d x = \int - d \frac{v}{v} = - ln v = - ln cos x$ $intsinx/cosxdx=int-dv/v=-lnv=-lncosx$

Hence $\int ({sec}^{2} x - 1) (\frac{sin x}{cos x}) d x = \frac{{tan}^{2} x}{2} + ln cos x + c$ $int(sec^2x-1)(sinx/cosx)dx=tan^2x/2+lncosx+c$

How do you integrate (sec(x)^2-1)(sin(x)/cos(x))(sec(x)2−1)(sin(x)cos(x))(sec(x)^2-1)(sin(x)/cos(x))?