Given to solve ,
`int 4/(csc(theta)-cot(theta)) d theta`
just for the convenience let` x= theta`
so,` int 4/(csc(theta)-cot(theta)) d theta`
=`int 4/(csc(x)-cot(x)) d x`
=`int 4/((1/sin(x))-(cos(x)/sin(x))) d x`
=`int 4/((1-cosx)/sinx) dx`
= `int 4sinx/(1-cosx) dx`----------(1)
let `u = 1-cosx`
=> `du=sinx dx`
so , the equation (1) implies
`int 4sinx/(1-cosx) dx`
= `4 int (1/u) du`
= `4 ln(u)+c`
= `4ln(1-cosx)+c`
but `x= theta` , so
= `4ln(1-costheta)+c`
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