Question

Question #f550a

Answer 1

`int\ (1-sin^2(x))/sin^2(x)\ dx=-cot(x)-x+C`

Explanation:

We can first split the fraction into two:
`int\ (1-sin^2(x))/sin^2(x)\ dx=int\ 1/sin^2(x)-sin^2(x)/sin^2(x)\ dx=`

`=int\ 1/sin^2(x)-1\ dx=int\ 1/sin^2(x)\ dx-x`

We can now use the following identity:
`1/sin(theta)=csc(theta)`

`int\ csc^2(x)\ dx-x`

We know that the derivative of `cot(x)` is `-csc^2(x)`, so we can add a minus sign both outside and inside the integral (so they cancel) to work it out:
`-int\ -csc^2(x)\ dx-x=-cot(x)-x+C`