How do you find #\int [ 3\sin x - 2\sec ^ { 2} x ] d x#?

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Answer 1 · 2017-12-04T08:44:24

#int[3sinx-2sec^2x]dx=-3cosx-2tanx+C#

#int[3sinx-2sec^2x]dx#

Remembering that integration is the reverse of differentiation

and since

#d/(dx)(cosx)=-sinx#

and#d/(dx)(tanx)=sec^2x#

we have

#int[3sinx-2sec^2x]dx=-3cosx-2tanx+C#