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受欢迎的微积分 >

integral of sin^4(6x)cos^2(6x)

解答

\int sin^{4} (6 x) cos^{2} (6 x) d x

解答

\frac{1}{576} (36 x - 3 sin (12 x) - 4 sin^{3} (6 x) cos (6 x) + 16 sin^{5} (6 x) cos (6 x)) + C

求解步骤

\int sin^{4} (6 x) cos^{2} (6 x) d x

使用换元积分法

= \int sin^{4} (u) cos^{2} (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) cos^{2} (u) d u

使用三角恒等式改写

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

乘开

sin^{4} (u) (1 - sin^{2} (u)) : sin^{4} (u) - sin^{6} (u)

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

u = 6 x

代回

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

化简

\frac{1}{6} (- \frac{1}{4} sin^{3} (6 x) cos (6 x) + \frac{3}{8} (6 x - \frac{1}{2} sin (2 \cdot 6 x)) - (- \frac{cos ( 6 x ) sin ^{5} ( 6 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (6 x) cos (6 x) + \frac{3}{8} (6 x - \frac{1}{2} sin (2 \cdot 6 x))))) : \frac{1}{576} (36 x - 3 sin (12 x) - 4 sin^{3} (6 x) cos (6 x) + 16 sin^{5} (6 x) cos (6 x))

= \frac{1}{576} (36 x - 3 sin (12 x) - 4 sin^{3} (6 x) cos (6 x) + 16 sin^{5} (6 x) cos (6 x))

解答补常数

= \frac{1}{576} (36 x - 3 sin (12 x) - 4 sin^{3} (6 x) cos (6 x) + 16 sin^{5} (6 x) cos (6 x)) + C

作图

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