Can you do the reverse?

Calculus Level 3

∫ tan ⁡ x d x = ? \large \int \tan x \ dx = \ ?

Note that C C is an arbitrary constant.

1 2 tan ⁡ 2 x + C \frac 1 2 \tan^2 x + C sec ⁡ 2 x + C \sec^2 x + C ln ⁡ ∣ sec ⁡ x ∣ + C \ln \mid \sec x \mid + C ln ⁡ ∣ cos ⁡ x ∣ + C \ln \mid \cos x \mid + C sec ⁡ x tan ⁡ x + C \sec x \tan x + C

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Lew Sterling Jr by Lew Sterling Jr , 24, USA

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1 solution

Discussions for this problem are now closed

Raj Magesh
Dec 20, 2014

∫ tan ⁡ x d x = ∫ sin ⁡ x cos ⁡ x d x \int \tan{x} dx = \int \dfrac{\sin{x}}{\cos{x}}dx

Let t = cos ⁡ x t=\cos{x} . Then, d t = − sin ⁡ x d x dt=-\sin{x} dx or sin ⁡ x d x = − d t \sin{x} dx = -dt .

Substituting this into the integral above:

∫ − d t t = − ln ⁡ ∣ t ∣ + C = − ln ⁡ ∣ cos ⁡ x ∣ + C = ln ⁡ ∣ sec ⁡ x ∣ + C \int \dfrac{-dt}{t} = -\ln{|t|}+C = -\ln{|\cos{x}|} +C = \boxed{\ln{|\sec{x}|+C}}

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Moderator note:

Nice and simple. Do you know of any other alternative methods?

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