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#