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受欢迎的微积分 >

integral of sin^4(6θ)

解答

\int sin^{4} (6 θ) d θ

解答

\frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ))) + C

求解步骤

\int sin^{4} (6 θ) d θ

使用换元积分法

= \int sin^{4} (u) \frac{1}{6} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int sin^{4} (u) d u

使用降次积分法

= \frac{1}{6} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

\int sin^{2} (u) d u = \frac{1}{2} (u - \frac{1}{2} sin (2 u))

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

u = 6 θ

代回

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

化简

\frac{1}{6} (- \frac{cos ( 6 θ ) sin ^{3} ( 6 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (6 θ - \frac{1}{2} sin (2 \cdot 6 θ))) : \frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ)))

= \frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ)))

解答补常数

= \frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ))) + C

作图

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