Question

what is integration of cosec 2x dx

Answer 1

2x = t therefore 2dx = dt
therefore integration of cosec(2x)dx = 1/2cosec(t)dt=1/2ln|cosect - cott|= 1/2ln|cosec(2x)-cot(2x)|

Answer 2

To Find :

integration of cosec(2x)

Step-by-step explanation:

rewrite cosec -------> csc

$\displaystyle\int \csc (2 x)\,dx= \displaystyle\int \csc (2 x)\dfrac{\csc\,2\,x + \cot\,2\,x}{\csc\,2\,x + \cot\,2\,x}\,dx$

$= \displaystyle\int \dfrac{\csc^2\,2\,x + \csc\,2\,x\cot\,2\,x}{\csc\,2\,x + \cot\,2\,x}\,dx= \displaystyle\int \dfrac{ \csc\,2\,x\cot\,2\,x + \csc^2\,2\,x }{\csc\,2\,x + \cot\,2\,x}\,dx$

$= - \dfrac{1}{2} \displaystyle\int \dfrac{ - 2\csc\,2\,x\cot\,2\,x - 2 \csc^2\,2\,x }{\csc\,2\,x + \cot\,2\,x}\,dx=-\dfrac{1}{2} \ln |\csc\,2\,x + \cot\,2\,x| + c=-\dfrac{1}{2} \ln |\dfrac{1}{\sin\,2\,x} +\dfrac{ \cos\,2\,x}{\sin\,2\,x}| + c=-\dfrac{1}{2} \ln |\dfrac{1 + \cos\,2\,x}{\sin\,2\,x} | + c=-\dfrac{1}{2} \ln |\dfrac{2\cos^2\,x}{2\sin\,x\cos\,x} | + c=-\dfrac{1}{2} \ln |\cot\,x | + c$

Hence :

$\displaystyle\int \csc (2 x)\,dx = -\dfrac{1}{2} \ln |\cot\,x | + c$    

#SPJ2