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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


Sagot :

Answer:

[tex]\int\limits_c \sec^2 z dz = i\tanh(\frac{\pi}{4}) -1[/tex]

Step-by-step explanation:

Given

[tex]\int\limits_c \sec^2 z dz[/tex]

From:

[tex]\frac{\pi}{4}[/tex] to [tex]\frac{\pi i}{4}[/tex]

Required

Integrate by first method

Let:

[tex]f(z) = \sec^2z[/tex] and [tex]F(z) = \tan z[/tex]

[tex]\int\limits_c \sec^2 z dz[/tex] from [tex]\frac{\pi}{4}[/tex] to [tex]\frac{\pi i}{4}[/tex] implies that:

[tex]\int\limits_c \sec^2 z dz = F(\frac{\pi}{4}i) - F(\frac{\pi}{4})[/tex]

Recall that:

[tex]F(z) = \tan z[/tex]

So:

[tex]F(\frac{\pi}{4}i) = \tan(\frac{\pi}{4}i)[/tex]

[tex]F(\frac{\pi}{4}) = \tan(\frac{\pi}{4})[/tex]

So, we have:

[tex]\int\limits_c \sec^2 z dz = F(\frac{\pi}{4}i) - F(\frac{\pi}{4})[/tex]

[tex]\int\limits_c \sec^2 z dz = \tan(\frac{\pi}{4}i) -\tan(\frac{\pi}{4})[/tex]

In trigonometry:

[tex]\tan(\frac{\pi}{4}) = 1[/tex]

and

[tex]\tan(\frac{\pi}{4}i) = i\tanh(\frac{\pi}{4})[/tex]

So:

[tex]\int\limits_c \sec^2 z dz = i\tanh(\frac{\pi}{4}) -1[/tex]