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受欢迎的微积分 >

integral of sin^5(9x)

解答

\int sin^{5} (9 x) d x

解答

\frac{1}{9} (- cos (9 x) + \frac{2 cos ^{3} ( 9 x )}{3} - \frac{cos ^{5} ( 9 x )}{5}) + C

求解步骤

\int sin^{5} (9 x) d x

使用换元积分法

= \int sin^{5} (u) \frac{1}{9} d u

提出常数:

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

= \frac{1}{9} \cdot \int sin^{5} (u) d u

使用三角恒等式改写

= \frac{1}{9} \cdot \int (1 - cos^{2} (u))^{2} sin (u) d u

使用换元积分法

= \frac{1}{9} \cdot \int - 1 + 2 v^{2} - v^{4} 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}{9} (- \int 1 d v + \int 2 v^{2} d v - \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

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

= \frac{1}{9} (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

代回

= \frac{1}{9} (- cos (9 x) + \frac{2 cos ^{3} ( 9 x )}{3} - \frac{cos ^{5} ( 9 x )}{5})

解答补常数

= \frac{1}{9} (- cos (9 x) + \frac{2 cos ^{3} ( 9 x )}{3} - \frac{cos ^{5} ( 9 x )}{5}) + C

作图

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