integral of sin^4(5x)

解答

\int sin^{4} (5 x) d x

\frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x))) + C

求解步骤

\int sin^{4} (5 x) d x

使用换元积分法

= \int sin^{4} (u) \frac{1}{5} d u

提出常数:

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

= \frac{1}{5} \cdot \int sin^{4} (u) d u

使用降次积分法

= \frac{1}{5} (- \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}{5} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

u = 5 x

代回

= \frac{1}{5} (- \frac{cos ( 5 x ) sin ^{3} ( 5 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (5 x - \frac{1}{2} sin (2 \cdot 5 x)))

化简

\frac{1}{5} (- \frac{cos ( 5 x ) sin ^{3} ( 5 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (5 x - \frac{1}{2} sin (2 \cdot 5 x))) : \frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x)))

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x)))

解答补常数

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (10 x))) + C