Question

$\displaystyle \sf \: \int \: \frac{ \sec(x) {}^{2} - 1}{ \cot(x) }$
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Answer 1

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

Given integral is

$\rm \: \displaystyle\int\rm \dfrac{ {sec}^{2}x - 1 }{cotx} \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{ {tan}^{2}x }{cotx} \: dx$

$\rm \: = \: \displaystyle\int\rm {tan}^{3}x \: dx$

$\rm \: = \: \displaystyle\int\rm tanx \times {tan}^{2}x \: dx$

$\rm \: = \: \displaystyle\int\rm tanx \times ( {sec}^{2}x - 1) \: dx$

$\rm \: = \: \displaystyle\int\rm tanx \: {sec}^{2}x \: dx \: - \: \displaystyle\int\rm tanx \: dx$

In the first integral, we use method of Substitution.

So, Substitute

$\rm \: tanx = y$

$\rm \: {sec}^{2} x \: dx=d y$

$\rm \: = \: \displaystyle\int\rm y \: dy \: - \: log |secx| + c$

$\rm \: = \: \frac{ {y}^{2} }{2} \: - \: log |secx| + c$

$\rm \: = \: \frac{ {tan}^{2} x}{2} \: - \: log |secx| + c$

Hence,

$\boxed{ \rm{ \:\rm \:\displaystyle\int\rm \frac{ {sec}^{2}x - 1 }{cotx}dx = \: \frac{ {tan}^{2} x}{2} \: - \: log |secx| + c \: \: }}$

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Formulae Used :-

$\boxed{ \rm{ \: {sec}^{2}x - {tan}^{2}x = 1 \: }}$

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

$\boxed{ \rm{ \:\displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

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Additional Information :-

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

Answer 2

Answer:

$\displaystyle \sf \: \int \: \frac{ \sec(x) {}^{2} - 1}{ \cot(x) } dx$

$\large\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

Take the integral:

$\\ \rm\[ \int \tan (x)\left(\sec ^{2}(x)-1\right) d x \]$

Expanding the integrand $\rm \tan (x)\left(\sec ^{2}(x)-1\right)$ gives $\rm\tan (x) \sec ^{2}(x)-\tan (x) :$

$\\ \rm\[ =\int\left(\tan (x) \sec ^{2}(x)-\tan (x)\right) d x \]$

Integrate the sum term by term and factor out constants:

$\\ \rm\[ =\int \tan (x) \sec ^{2}(x) d x-\int \tan (x) d x \]$

For the integrand $\rm \tan (x) \sec ^{2}(x), \operatorname{substitute} u=\tan (x)$ and $\rm d u=\sec ^{2}(x) d x$ :

$\\ \rm =\int u \: d u-\int \tan (x) d x$

$\\ \rm\[ =\frac{u^{2}}{2}-\int \tan (x) d x \]$

$\text{Rewrite \( \rm\tan (x) \) as \( \rm \dfrac{\sin (x)}{\cos (x)} \)}$

$\\ \rm \[ =\frac{u^{2}}{2}-\int \frac{\sin (x)}{\cos (x)} d x \]$

$\text{For the integrand \( \rm\dfrac{\sin (x)}{\cos (x)} \), substitute \( \rm s=\cos (x) \) and \( \rm d s=-\sin (x) d x: \)}$

$\\ \rm \[ =\frac{u^{2}}{2}-\int-\frac{1}{s} d s \]$

Factor out constants:

$\\ \rm\[ =\frac{u^{2}}{2}+\int \frac{1}{s} d s \]$

$\text{The integral of \( \rm\dfrac{1}{s} \) is \( \rm \log (s): \)}$

$\\ \rm\[ =\log (s)+\frac{u^{2}}{2}+\text { c } \]$

$\text{Substitute back for \( \rm s=\cos (x): \)}$

$\\ \rm =\frac{u^{2}}{2}+\log (\cos (x))+ c$

$\text{Substitute back for \( \rm u=\tan (x) \) :}$

$\\ \rm \[ =\frac{\tan ^{2}(x)}{2}+\log (\cos (x))+\text { c } \]$

Which is equivalent for restricted x values to:

Answer: $\color{purple} \boxed{ \displaystyle\rm\[ \frac{\sec ^{2}(x)}{2}+\log (\cos (x))+\text { c } \]}$