integral of (tan(5w)+cot(5w))^2

解

\int (tan (5 w) + cot (5 w))^{2} d w

\frac{1}{5} tan (5 w) - \frac{1}{5} cot (5 w) + C

解答ステップ

\int (tan (5 w) + cot (5 w))^{2} d w

拡張

(tan (5 w) + cot (5 w))^{2} : tan^{2} (5 w) + 2 tan (5 w) cot (5 w) + cot^{2} (5 w)

= \int tan^{2} (5 w) + 2 tan (5 w) cot (5 w) + cot^{2} (5 w) d w

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int tan^{2} (5 w) d w + \int 2 tan (5 w) cot (5 w) d w + \int cot^{2} (5 w) d w

\int tan^{2} (5 w) d w = - w + \frac{1}{5} tan (5 w)

\int 2 tan (5 w) cot (5 w) d w = 2 w

\int cot^{2} (5 w) d w = - w - \frac{1}{5} cot (5 w)

= - w + \frac{1}{5} tan (5 w) + 2 w - w - \frac{1}{5} cot (5 w)

簡素化

= \frac{1}{5} tan (5 w) - \frac{1}{5} cot (5 w)

定数を解答に追加する

= \frac{1}{5} tan (5 w) - \frac{1}{5} cot (5 w) + C