Question

$\huge{\boxed{ \bold{ \red{{ \int\csc(x) \sec(x) \: dx}}}}}$






❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​

Answer 1

Step-by-step explanation:

given :

• $\huge{\boxed{ \bold{ \red{{ \int\csc(x) \sec(x) \: dx}}}}}$

to find :

• sec(x) \: dx}}}}}[/tex]

solution :

• Note: Here, we are" considering " log x as log_ex`.

• Let I = {sec x "cosec "x}/{log ( tan x)} dx

• "Putting" "log" \ tan x = t`

• \[\Rightarrow \frac{\sec^2 x}{\tan x} =

• \frac{dt}{dx}\]

• \[\Rightarrow \text{sec x cosec x dx} =

• dt\]

• \[\therefore I = \int\frac{1}{t}dt\]

• \[= \text{log }\left| \text{t}\right| + C\] \[= \text{log} \left| \text{log} \left( \tan

• x \right) \right| + C\]

Answer 2

$\int{\csc{\left(x \right)} \sec{\left(x \right)} }d x$

$\color{red}{\int{\csc{\left(x \right)} \sec{\left(x \right)} d x}} = \color{red}{\int{\frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}}$

$\color{red}{\int{\frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}} = \color{red}{\int{\frac{\sin{\left(x \right)}}{\left(1 - \cos^{2}{\left(x \right)}\right) \cos{\left(x \right)}} d x}}$

$Let \: u=\cos{\left(x \right)}$

$\color{red}{\int{\frac{\sin{\left(x \right)}}{\left(1 - \cos^{2}{\left(x \right)}\right) \cos{\left(x \right)}} d x}} = \color{red}{\int{\left(- \frac{1}{u \left(1 - u^{2}\right)}\right)d u}}$

$\color{red}{\int{\left(- \frac{1}{u \left(1 - u^{2}\right)}\right)d u}} = \color{red}{\left(- \int{\frac{1}{u \left(1 - u^{2}\right)} d u}\right)}$

$- \color{red}{\int{\frac{1}{u \left(1 - u^{2}\right)} d u}} = - \color{red}{\int{\left(- \frac{1}{2 \left(u + 1\right)} - \frac{1}{2 \left(u - 1\right)} + \frac{1}{u}\right)d u}}$

$- \color{red}{\int{\left(- \frac{1}{2 \left(u + 1\right)} - \frac{1}{2 \left(u - 1\right)} + \frac{1}{u}\right)d u}} = - \color{red}{\left(\int{\frac{1}{u} d u} - \int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u}\right)}$

$\int{\frac{1}{2 \left(u - 1\right)} d u} + \int{\frac{1}{2 \left(u + 1\right)} d u} - \color{red}{\int{\frac{1}{u} d u}} = \int{\frac{1}{2 \left(u - 1\right)} d u} + \int{\frac{1}{2 \left(u + 1\right)} d u} - \color{red}{\ln{\left(u \right)}}$

$- \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \color{red}{\int{\frac{1}{2 \left(u + 1\right)} d u}} = - \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \color{red}{\left(\frac{\int{\frac{1}{u + 1} d u}}{2}\right)}$

Let v=u + 1.

$- \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{\color{red}{\int{\frac{1}{u + 1} d u}}}{2} = - \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{\color{red}{\int{\frac{1}{v} d v}}}{2}$

$- \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{\color{red}{\int{\frac{1}{v} d v}}}{2} = - \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{\color{red}{\ln{\left(v \right)}}}{2}$

$- \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{\ln{\left(\color{red}{v} \right)}}{2} = - \ln{\left(u \right)} + \int{\frac{1}{2 \left(u - 1\right)} d u} + \frac{\ln{\left(\color{red}{\left(u + 1\right)} \right)}}{2}$

$- \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \color{red}{\int{\frac{1}{2 \left(u - 1\right)} d u}} = - \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \color{red}{\left(\frac{\int{\frac{1}{u - 1} d u}}{2}\right)}$

Let v=u - 1

$- \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \frac{\color{red}{\int{\frac{1}{u - 1} d u}}}{2} = - \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \frac{\color{red}{\int{\frac{1}{v} d v}}}{2}$

$- \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \frac{\color{red}{\int{\frac{1}{v} d v}}}{2} = - \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \frac{\color{red}{\ln{\left(v \right)}}}{2}$

$- \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \frac{\ln{\left(\color{red}{v} \right)}}{2} = - \ln{\left(u \right)} + \frac{\ln{\left(u + 1 \right)}}{2} + \frac{\ln{\left(\color{red}{\left(u - 1\right)} \right)}}{2}$

$\frac{\ln{\left(-1 + \color{red}{u} \right)}}{2} + \frac{\ln{\left(1 + \color{red}{u} \right)}}{2} - \ln{\left(\color{red}{u} \right)} = \frac{\ln{\left(-1 + \color{red}{\cos{\left(x \right)}} \right)}}{2} + \frac{\ln{\left(1 + \color{red}{\cos{\left(x \right)}} \right)}}{2} - \ln{\left(\color{red}{\cos{\left(x \right)}} \right)}$

Therefore,

$\int{\csc{\left(x \right)} \sec{\left(x \right)} d x} = \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$

Answer:

$\int{\csc{\left(x \right)} \sec{\left(x \right)} d x}=\frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$