Question

integrate cosec(ax+b) ​

Answer 1

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

$\rm \: \displaystyle\int\rm cosec(ax + b) \: dx$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm \: ax + b = y$

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

$\rm \: \dfrac{d}{dx}(ax + b) = \dfrac{d}{dx}y$

$\rm \: a = \dfrac{dy}{dx}$

$\rm\implies \:dx \: = \: \dfrac{1}{a} \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm cosecy \times \dfrac{1}{a} \: dy$

$\rm \: = \: \dfrac{1}{a} \: ( - \: cosecy \: coty \: ) \: + \: c$

$\rm \: = \: - \: \dfrac{cosecy \: coty}{a} \: + \: c$

$\rm \: = \: - \: \dfrac{cosec(ax + b) \: cot(a + b)}{a} \: + \: c$

Hence

$\rm \:\displaystyle\int\rm cosec(ax + b) \: dx = \: - \: \dfrac{cosec(ax + b) \: cot(a + b)}{a} \: + \: c$

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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

refer the given attachment