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受欢迎的微积分 >

integral of 9sin^2(xco)s^3x

解答

\int 9 sin^{2} (x co) s^{3} x d x

解答

\frac{9 s ^{3}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

求解步骤

\int 9 sin^{2} (x co) s^{3} x d x

提出常数:

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

= 9 s^{3} \cdot \int sin^{2} (co x) x d x

使用换元积分法

= 9 s^{3} \cdot \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

提出常数:

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

= 9 s^{3} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

使用分布积分法

= 9 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u)

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

= 9 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u)))

u = co x

代回

= 9 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x)))

化简

9 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x))) : \frac{9 s ^{3}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

= \frac{9 s ^{3}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

解答补常数

= \frac{9 s ^{3}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

作图

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