Recall from the previous section Work by a Variable Force, that:

`"Work"=intFdx`

So we need to calculate

`W=int(2+tan\ x)/(cos x)dx`

Since

`(2+tan\ x)/(cos x) =2/(cos x) +(tan\ x)/(cos x)`

`=2\ sec\ x+tan\ x\ sec\ x`,

we have, using the given formulas:

`W =int(2\ sec\ x+tan\ x\ sec\ x)dx`

`=2\ ln\|sec\ x+tan\ x|+sec\ x+K`

Since `W = 0` when `x = 0`, we have:

`0=2\ ln\ |sec\ 0+` `tan\ 0|+sec\ 0+K`

`0=2(0)+1+K`

So `K = -1`. So we have:

`W=2\ ln\ |sec\ x+tan\ x|+` `sec\ x-1`

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