Question

${ \boxed{ \bold \red{ \int \cot(ab) db}}}$



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

Answer 1

Answer:

1. Proof

Strategy: Make in terms of sin's and cos's; Use substitution.

(integral) cot x dx = (integral) cos x

sin x dx

set

u = sin x.

then we find

du = cos x dx

substitute du=cos x, u=sin x

(integral) cos x

sin x dx = (integral)

du

u

solve integral

= ln |u| + C

substitute back u=sin x

= ln |sin x| + C

Answer 2

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

Given integral is

$\red{\rm :\longmapsto\: \displaystyle\int \:cot(ab) \: db }$

can be rewritten as

$\rm \: = \: \displaystyle\int \: \frac{cos(ab)}{sin(ab)} \: db$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:sin(ab) = x}$

On differentiating both sides w. r. t. b, we get

$\red{\rm :\longmapsto\:a \: cos(ab) \: db= dx}$

$\red{\rm :\longmapsto\: \: cos(ab) \: db= \dfrac{dx}{a} }$

So, on substituting these values, we get

$\rm \: = \: \displaystyle\int \: \frac{dx}{a \: x}$

$\rm \: = \:\dfrac{1}{a} \: \displaystyle\int \: \frac{dx}{\: x \: }$

$\rm \: = \:\dfrac{1}{a} \: logx \: + \: c$

$\rm \: = \:\dfrac{1}{a} \: logsin(ab)\: + \: c$

Hence,

$\purple{\rm \implies\:\boxed{ \tt{ \: \displaystyle\int \:cot(ab) \: db = \:\dfrac{1}{a} \: logsin(ab)\: + \: c \: }}}$

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Additional Information :-

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