Question

Evaluate

$\bigstar \underline{ \boxed{\displaystyle{ \sf{ \red{ \int sec \: x(sec \: x+tan \: x) dx}}} }} \bigstar$
​

Answer 1

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

Given integral is

$\displaystyle{ \sf{ \red{ \rm :\longmapsto\:\int sec \: x(sec \: x+tan \: x) dx}}}$

$\rm \:  =  \: \displaystyle\int\rm [ \: {sec}^{2}x \: + \: secx \: tanx \: ] \: dx$

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

$\rm \:  =  \: tanx \: + \: secx \: + \: c$

Hence,

$\boxed{ \tt{ \: \displaystyle{ \sf{ \red{ \rm \:\int secx(secx+tan x) dx \: = \:tanx + secx + c}}}}}$

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More to Know :-

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

Answer 2

Question:-

$\sf \int sec \: x(sec \: x + tan \: x)dx$

Given:-

$\sf I = \int sec \: x(sec \: x + tan \: x)dx$

To Find:

● Evaluate:-

$\sf I = \int sec \: x(sec \: x + tan \: x)dx$

Solution:-

We know that,

$\sf I = \int sec \: x(sec \: x + tan \: x)dx$

$\longmapsto\sf I = \int \: sec {}^{2} x \: dx + \int \: sec x \: tanx \: dx$

$\longmapsto\sf I = \int tan \: x \: + sec x \: + c$

Hence,Solved.

Answer:-

$\underline{ \boxed{\displaystyle{ \sf{ \red{ \int sec \: x(sec \: x+tan \: x) dx = tanx \: + secx \: + c}}} }}$