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

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Answer 1 · 2016-07-03T03:36:22

                  #=-2*csc^2(x)*cot(x)#

#1/sin^2(x) = csc^2(x)#

Treat #csc^2(x)# as #(csc(x))^2#.

Using power rule, bring down 2 as coefficient.

That would give us #2*csc(x)#

Don't forget chain rule and multiply with the derivative of #csc(x)#.

We know that the derivative of #csc(x)# is #-cot(x)*csc(x)#.

#d/dx csc^2(x)=-2*csc(x)*cot(x)*csc(x)#
#=-2*csc^2(x)*cot(x)#