How do you find $int sinx -1/csc(x)dx$ ?

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Answer 1 · 2018-06-12T00:41:02

$C$ where $C$ is an arbitrary constant.

Recall that $cscx=1/sinx.$ Rewrite the integral:

$int(sinx-1/cscx)dx=int(sinx-1/(1/sinx))dx$

$=int(sinx-sinx)dx$

$=int(0)dx=C$ where $C$ is an arbitrary constant.

We can easily see this is true as the derivative of any constant $C$ , or $d/dx(C),$ is $0.$

Answer 2 · 2018-06-12T14:22:35

$C$ , in which $C$ is a constant.

Given: $intsinx-1/cscx \ dx$ .

Notice how 1/cscx=sinx#, so the integral becomes:

$=intsinx-sinx \ dx$

$=int0 \ dx$

$=C$

How do you find int sinx -1/csc(x)dx int sinx -1/csc(x)dx ?