How do you find #int cosxcscxdx #?

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Answer 1 · 2018-04-10T22:05:38

#intcosxcscxdx=ln|sinx|+C#

Recall that #cscx=1/sinx,# giving us

#intcosx/sinxdx#.

This can be solved with a substitution.

#u=sinx#

#du=cosxdx#

We then have

#int(du)/u=ln|u|+C#

Rewriting in terms of #x# yields

#intcosxcscxdx=ln|sinx|+C#