Math Is Fun Forum

Here another way to it
∫▒〖1/sin^2⁡x  dx〗
let t=tan⁡〖x/2〗
dt/dx=1/2  sec^2⁡〖x/2〗
therefore dx=2dtcos^2  x/2
now tan x/2=t
thefore cos x/2=1/√(1+t^2 )   
sin⁡〖x/2〗=t/√(1+t^2 )
now sinx=2sin x/2 cos x/2
hence sinx=2t/(1+t^2 )  therefore sin^2⁡〖x=4t^2/(1+t^2 )^2 〗
now substituting x by t we get 
∫▒1/sin^2⁡x  dx= ∫▒(1+t^2 )^2/(4t^2 )×2/((1+t^2)) dt
=1/2 ∫▒(1+t^2)/t^2   dt
=1/2 ∫▒1/t^2 +1 dt 
=1/2  [(-1 )/t+t]+c
=-[(1-t^2)/2t]+c
=-[(1-tan^2⁡〖x/2〗)/(2tan x/2)]+c
now tanx=(2tan x/2)/(1-tan^2⁡〖x/2〗 )    hence cotx=(1-tan^2⁡〖x/2〗)/(2tan x/2)
therefore ∫▒〖1/〖sin〗^2⁡x  dx〗=-cotx+c