Question

indefinite integration cosec²(3x+2)dx​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int cosec^{2} (3x +2 ) \: dx$

Let us assume that:

$\displaystyle \rm \longrightarrow u = 3x + 2$

$\displaystyle \rm \longrightarrow du = 3 \: dx$

$\displaystyle \rm \longrightarrow dx = \dfrac{du}{3}$

Therefore, we have:

$\displaystyle \rm \longrightarrow I = \int \dfrac{1}{3} cosec^{2} (u) \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{3} \int cosec^{2} (u) \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{3} \times - \cot(u) + C$

$\displaystyle \rm \longrightarrow I = \dfrac{ - \cot(u) }{3}+ C$

Substituting the value of u, we get:

$\displaystyle \rm \longrightarrow I = \dfrac{ - \cot(3x +2 ) }{3}+ C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

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

To Evaluate :

$\displaystyle \rm \longrightarrow I = \int cosec^{2} (3x +2 )dx$

Putting 3x+2=t

${ \displaystyle{ \rm{ \implies \: 3dx = dt}}}$

${ \displaystyle{ \rm{ \implies \: dx = \frac{dt}{3} }}}$

${ \displaystyle{ \rm{ \implies \: \therefore \: \int \: \cosec {}^{2} t \frac{dt}{3} }}}$

${ \displaystyle{ \rm{ \implies \: = - \frac{ 1}{3} \cot \: t \: + c }}}$

$\bigstar { \displaystyle{ \rm{ \implies \: = - \frac{ 1}{3} \cot \: (3x + 2) \: + c }}}$

${ \underline{ \underline{\rule{200pt}{5pt}}}}$