integral of sin^4(3x)-cos^4(3x)

解

\int sin^{4} (3 x) - cos^{4} (3 x) d x

x sin^{4} (3 x) - x cos^{4} (3 x) + 2 x cos^{2} (3 x) - x - \frac{1}{6} sin (6 x) + C

解答ステップ

\int sin^{4} (3 x) - cos^{4} (3 x) d x

部分積分を適用する

= (sin^{4} (3 x) - cos^{4} (3 x)) x - \int 12 cos (3 x) sin (3 x) x d x

\int 12 cos (3 x) sin (3 x) x d x = - 2 x cos^{2} (3 x) + x + \frac{1}{6} sin (6 x)

= (sin^{4} (3 x) - cos^{4} (3 x)) x - (- 2 x cos^{2} (3 x) + x + \frac{1}{6} sin (6 x))

簡素化

(sin^{4} (3 x) - cos^{4} (3 x)) x - (- 2 x cos^{2} (3 x) + x + \frac{1}{6} sin (6 x)) : x sin^{4} (3 x) - x cos^{4} (3 x) + 2 x cos^{2} (3 x) - x - \frac{1}{6} sin (6 x)

= x sin^{4} (3 x) - x cos^{4} (3 x) + 2 x cos^{2} (3 x) - x - \frac{1}{6} sin (6 x)

定数を解答に追加する

= x sin^{4} (3 x) - x cos^{4} (3 x) + 2 x cos^{2} (3 x) - x - \frac{1}{6} sin (6 x) + C