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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x)) + C

求解步骤

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

使用换元积分法

= \int cos^{2} (u) 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

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

使用三角恒等式改写

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

乘开

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

= \frac{1}{2} \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}{2} (\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}{2} (- \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 = 2 x

代回

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

化简

\frac{1}{2} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) - (- \frac{cos ( 2 x ) sin ^{5} ( 2 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (2 \cdot 2 x))))) : \frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x))

= \frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x))

解答补常数

= \frac{1}{192} (12 x - 3 sin (4 x) - 4 sin^{3} (2 x) cos (2 x) + 16 sin^{5} (2 x) cos (2 x)) + C

作图

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