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受欢迎的微积分 >

integral of 32sin^4(2x)

解答

\int 32 sin^{4} (2 x) d x

解答

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

求解步骤

\int 32 sin^{4} (2 x) d x

提出常数:

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

= 32 \cdot \int sin^{4} (2 x) d x

使用换元积分法

= 32 \cdot \int sin^{4} (u) \frac{1}{2} d u

提出常数:

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

= 32 \cdot \frac{1}{2} \cdot \int sin^{4} (u) d u

使用降次积分法

= 32 \cdot \frac{1}{2} (- \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))

= 32 \cdot \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

u = 2 x

代回

= 32 \cdot \frac{1}{2} (- \frac{cos ( 2 x ) sin ^{3} ( 2 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 x - \frac{1}{2} sin (2 \cdot 2 x)))

化简

32 \cdot \frac{1}{2} (- \frac{cos ( 2 x ) sin ^{3} ( 2 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 x - \frac{1}{2} sin (2 \cdot 2 x))) : 16 (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x)))

= 16 (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x)))

解答补常数

= 16 (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x))) + C

作图

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