Question

Integrate by the first method or state why it does not apply and use the second method. Show the details.

â«sec²z dz, any path from Ï/4 to Ïi/4

Answer 1

`\int\limits_c \sec^2 z dz = i\tanh((\pi)/(4)) -1`

Explanation:

Given

`\int\limits_c \sec^2 z dz`

From:

`(\pi)/(4)` to
`(\pi i)/(4)`

Required

Integrate by first method

Let:

`f(z) = \sec^2z` and
`F(z) = \tan z`

`\int\limits_c \sec^2 z dz` from
`(\pi)/(4)` to
`(\pi i)/(4)` implies that:

`\int\limits_c \sec^2 z dz = F((\pi)/(4)i) - F((\pi)/(4))`

Recall that:

`F(z) = \tan z`

So:

`F((\pi)/(4)i) = \tan((\pi)/(4)i)`

`F((\pi)/(4)) = \tan((\pi)/(4))`

So, we have:

`\int\limits_c \sec^2 z dz = F((\pi)/(4)i) - F((\pi)/(4))`

`\int\limits_c \sec^2 z dz = \tan((\pi)/(4)i) -\tan((\pi)/(4))`

In trigonometry:

`\tan((\pi)/(4)) = 1`

and

`\tan((\pi)/(4)i) = i\tanh((\pi)/(4))`

So:

`\int\limits_c \sec^2 z dz = i\tanh((\pi)/(4)) -1`