integral of sin^4(4θ)

解答

\int sin^{4} (4 θ) d θ

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

求解步骤

\int sin^{4} (4 θ) d θ

使用换元积分法

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

提出常数:

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

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

使用降次积分法

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

u = 4 θ

代回

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

化简

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

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

解答补常数

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

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