d-1-2-tan-2cot-2-d-2tan-2-1-2-cot-

Question Number 37432 by MJS last updated on 13/Jun/18

\int \frac{d α}{\frac{1}{2} \tan α + 2 \cot \frac{α}{2}} = ?

\int \frac{d β}{2 \tan \frac{β}{2} + \frac{1}{2} \cot β} = ?

Answered by ajfour last updated on 13/Jun/18

\int \frac{2 \sin α \cos α d α}{\sin^{2} α + 4 \cos^{2} α} = \int \frac{\sin 2 α d α}{1 + \frac{3}{2} (1 + \cos 2 α)}

= \frac{1}{3} \int \frac{3 \sin 2 α d (2 α)}{5 + 3 \cos 2 α}

= - \frac{1}{3} \ln (5 + 3 \cos 2 α) + c .

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18

2) t = t a n \frac{β}{2} d t = {s e c}^{2} \frac{β}{2} \times \frac{1}{2} \times d β

\int \frac{2 d t}{1 + t^{2}} \times \frac{1}{2 t + \frac{1}{2} \times \frac{1 - t^{2}}{2 t}}

\int \frac{2 d t}{1 + t^{2}} \times \frac{4 t}{8 t^{2} + 1 - t^{2}}

8 \int \frac{t d t}{1 + t^{2}} \times \frac{1}{1 + 7 t^{2}} k = t^{2} d k = 2 t d t

4 \int \frac{d k}{1 + k} \times \frac{1}{1 + 7 k}

\frac{4}{6} \int \frac{(7 + 7 k) - (1 + 7 k)}{(1 + k) (1 + 7 k)} d k

\frac{2}{3} \int \frac{7 d k}{1 + 7 k} - \frac{2}{3} \int \frac{d k}{1 + k}

\frac{2}{3} l n ∣ \frac{1 + 7 k}{1 + k} ∣

\frac{2}{3} l n ∣ \frac{1 + 7 t^{2}}{1 + t^{2}} ∣

\frac{2}{3} l n ∣ \frac{1 + 7 {t a n}^{2} β}{1 + {t a n}^{2} β} ∣