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受欢迎的微积分 >

integral of sin^4(2x)cos^3(2x)

解答

\int sin^{4} (2 x) cos^{3} (2 x) d x

解答

\frac{1}{2} (\frac{sin ^{5} ( 2 x )}{5} - \frac{sin ^{7} ( 2 x )}{7}) + C

求解步骤

\int sin^{4} (2 x) cos^{3} (2 x) d x

使用换元积分法

= \int sin^{4} (u) cos^{3} (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 sin^{4} (u) cos^{3} (u) d u

使用三角恒等式改写

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

使用换元积分法

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

乘开

v^{4} (1 - v^{2}) : v^{4} - v^{6}

= \frac{1}{2} \cdot \int v^{4} - v^{6} d v

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 v^{4} d v - \int v^{6} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{2} (\frac{v ^{5}}{5} - \frac{v ^{7}}{7})

代回

= \frac{1}{2} (\frac{sin ^{5} ( 2 x )}{5} - \frac{sin ^{7} ( 2 x )}{7})

解答补常数

= \frac{1}{2} (\frac{sin ^{5} ( 2 x )}{5} - \frac{sin ^{7} ( 2 x )}{7}) + C

作图

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