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

```
#=-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 2`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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