integral of sin^6(θ)

解答

\int sin^{6} (θ) d θ

- \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ))) + C

求解步骤

\int sin^{6} (θ) d θ

使用降次积分法

= - \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} \cdot \int sin^{4} (θ) d θ

\int sin^{4} (θ) d θ = - \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ))

= - \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ)))

解答补常数

= - \frac{cos ( θ ) sin ^{5} ( θ )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (θ) cos (θ) + \frac{3}{8} (θ - \frac{1}{2} sin (2 θ))) + C

x \to \infty lim (x^{2} - 3 x)

\frac{d}{d x} (50 sin (2000 x))

a re a x = y^{2} - 2, x = - y^{2} + 2

\int \frac{8 x}{x - 1} d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - \frac{25}{3}