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Question Number 179679 by ROOSEVELT last updated on 01/Nov/22

∫_2 ^4 tan (x)

$$\int_{\mathrm{2}} ^{\mathrm{4}} \mathrm{tan}\:\left({x}\right) \\ $$

Answered by Acem last updated on 01/Nov/22

a= −ln ∣cos x∣ ∣_2 ^4 = −ln ∣cos 4∣+ ln ∣cos 2∣ = 18.29 E10^(−4)

$${a}=\:−\mathrm{ln}\:\mid\mathrm{cos}\:{x}\mid\:\mid_{\mathrm{2}} ^{\mathrm{4}} \:=\:−\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{4}\mid+\:\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{2}\mid \\ $$$$\:=\:\mathrm{18}.\mathrm{29}\:{E}\mathrm{10}^{−\mathrm{4}} \\ $$

Commented by Frix last updated on 01/Nov/22

I don′t think angles are in degree when it comes to integrals −ln ∣cos 4∣ +ln ∣cos 2∣ ≈−.451524 if x is in degree: ∫tan x dx =−((180)/π)ln ∣cos x∣ and the answer would be ∫_(2°) ^(4°) tan x dx≈.104826

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{angles}\:\mathrm{are}\:\mathrm{in}\:\mathrm{degree}\:\mathrm{when}\:\mathrm{it} \\ $$$$\mathrm{comes}\:\mathrm{to}\:\mathrm{integrals} \\ $$$$−\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{4}\mid\:+\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{2}\mid\:\approx−.\mathrm{451524} \\ $$$$\mathrm{if}\:{x}\:\mathrm{is}\:\mathrm{in}\:\mathrm{degree}: \\ $$$$\int\mathrm{tan}\:{x}\:{dx}\:=−\frac{\mathrm{180}}{\pi}\mathrm{ln}\:\mid\mathrm{cos}\:{x}\mid \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{would}\:\mathrm{be} \\ $$$$\underset{\mathrm{2}°} {\overset{\mathrm{4}°} {\int}}\mathrm{tan}\:{x}\:{dx}\approx.\mathrm{104826} \\ $$

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