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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]