Question

Find the indefinite integral of the function using substitution method. integral sign cot(5x + 9)dx​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm cot(5x + 9) \: dx \:$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{cos(5x + 9)}{sin(5x + 9)} \: dx$

To solve this integral, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:sin(5x + 9) = y}$

$\red{\rm :\longmapsto\:cos(5x + 9) \times 5 \: dx \: = \: dy}$

$\red{\rm :\longmapsto\:cos(5x + 9) dx \: = \: \dfrac{1}{5} dy}$

So, above integral can be rewritten as

$\rm \:  =  \: \dfrac{1}{5}\displaystyle\int\rm \frac{dy}{y}$

$\rm \:  =  \: \dfrac{1}{5}logy + c$

$\rm \:  =  \: \dfrac{1}{5}log\bigg[sin(5x + 9)\bigg] + c$

Hence,

$\boxed{\tt{ \displaystyle\int\rm cot(5x + 9) \: dx =  \: \dfrac{1}{5}logsin(5x + 9) + c \: }}$

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Additional Information

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