How do you find the integral of # 1/sin^2(x)#?

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Answer 1 · 2016-05-27T18:06:04

Notice that

#dcotx/dx=d(cosx/sinx)/dx=[(cosx)'sinx-cosx*(sinx)']/[sin^2 x]= [-sin^2x-cos^2x]/[sin^2 x]=-1/[(sin^2 x)]#

Hence

#int 1/[sin^2x]dx=-cotx+c#

Answer 2 · 2016-06-04T14:19:18

Maybe more "intuitive" instead of remembering :

#int1/sin^2(x) dx=int (1/cos^2(x))/(sin^2(x)/cos^2(x))dx = int (1/cos^2(x))/tan^2(x)dx#

let's #u = tan(x)#

#du = 1/cos^2(x) dx#

#int 1/u^2du #

which is

#[-1/u]#

and remember that u = tan(x) ::

#[-1/tan(x)]#