Key to "Integrate ∫arctan⁡(2θ)dθ"

Isa Trevino

Answered question

2021-08-19

Integrate $\int \arctan (2 θ) d θ$

komunidadO

Skilled2021-08-20Added 86 answers

Take the integral: $\int \arctan (2 θ) d θ = \int \tan^{- 1} (2 θ) d θ$

$For the integrand \int \tan^{- 1} (2 θ) d θ, integrate by parts, \int f d g = f g - \int g d f, where$

$d f = \frac{2}{4 θ^{2} + 1} d θ, g = θ :$

$= θ \tan^{- 1} (2 θ) - \int \frac{2 θ}{4 θ^{2} + 1} d θ$

Factor out constants:

$= θ \tan^{- 1} (2 θ) - 2 \int \frac{θ}{4 θ^{2} + 1} d θ$

$For the integrand \frac{θ}{4 θ^{2} + 1}, substitute u = 4 θ^{2} + 1 and d u = 8 θ d θ :$

$= θ \tan^{- 1} (2 θ) - \frac{1}{4} \int \frac{1}{u} d u$

$The integral of \frac{1}{u} is \log (u) :$

$= θ \tan^{- 1} (2 θ) - \frac{\log (u)}{4} + C$

$Substitute back for u = 4 θ^{2} + 1 :$

$Answer := θ \tan^{- 1} - \frac{1}{4} \log (4 θ^{2} + 1) + C$

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