tan-tan-3-1-tan-3-d-

Question Number 152273 by peter frank last updated on 27/Aug/21

\int \frac{\tan θ + \tan^{3} θ}{1 + \tan^{3} θ} d θ

Answered by qaz last updated on 27/Aug/21

\int \frac{\tan θ + \tan^{3} θ}{1 + \tan^{3} θ} d θ

= \int \frac{\tan θ}{1 + \tan^{3} θ} d (\tan θ)

= \int \frac{xdx}{1 + x^{3}}

= \int \frac{x}{(x + 1) (x^{2} - x + 1)} dx

= \frac{1}{3} \int (\frac{x + 1}{x^{2} - x + 1} - \frac{1}{x + 1}) dx

= \frac{1}{6} \int \frac{2 x - 1 + 3}{x^{2} - x + 1} dx - \frac{1}{3} \int \frac{1}{x + 1} dx

= \frac{1}{6} \ln ∣ x^{2} - x + 1 ∣ + \frac{1}{6} \cdot \frac{1}{2 \cdot \sqrt{\frac{3}{4}}} \tan^{- 1} \frac{x - \frac{1}{2}}{\sqrt{\frac{3}{4}}} - \frac{1}{3} \ln ∣ x + 1 ∣ + C

= \frac{1}{6} \ln ∣ \frac{x^{2} - x + 1}{{(x + 1)}^{2}} ∣ + \frac{1}{6 \sqrt{3}} \tan^{- 1} \frac{2 x - 1}{\sqrt{3}} + C

= \frac{1}{6} \ln ∣ \frac{\tan^{2} θ - \tan θ + 1}{{(\tan θ + 1)}^{2}} ∣ + \frac{1}{6 \sqrt{3}} \tan^{- 1} \frac{2 \tan θ - 1}{\sqrt{3}} + C

Facebook Tweet Pin