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

Answer 1

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

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

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

Utilizing the power rule, decrease the coefficient by 2.

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)#

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Answer 2Evelyn Agard
To find the integral of ( \frac{1}{\sin^2(x)} ), you can use the trigonometric identity ( \frac{1}{\sin^2(x)} = \csc^2(x) ). Then, integrate ( \csc^2(x) ) with respect to ( x ), which results in ( -\cot(x) + C ), where ( C ) is the constant of integration. Therefore, the integral of ( \frac{1}{\sin^2(x)} ) is ( -\cot(x) + C ).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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