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受欢迎的微积分 >

integral of sin(x)cos(x)e^x

解答

\int sin (x) cos (x) e^{x} d x

解答

- \frac{e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{10} + C

求解步骤

\int sin (x) cos (x) e^{x} d x

使用三角恒等式改写

= \int \frac{1}{2} e^{x} sin (2 x) d x

提出常数:

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

= \frac{1}{2} \cdot \int e^{x} sin (2 x) d x

使用分布积分法

= \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - \int - \frac{1}{2} e^{x} cos (2 x) d x)

提出常数:

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

= \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - (- \frac{1}{2} \cdot \int e^{x} cos (2 x) d x))

使用分布积分法

= \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - (- \frac{1}{2} (e^{x} \frac{1}{2} sin (2 x) - \int e^{x} \frac{1}{2} sin (2 x) d x)))

因此

\int e^{x} \frac{1}{2} sin (2 x) d x = \frac{1}{2} (- \frac{1}{2} e^{x} cos (2 x) - (- \frac{1}{2} (e^{x} \frac{1}{2} sin (2 x) - \int e^{x} \frac{1}{2} sin (2 x) d x)))

整理

\int e^{x} \frac{1}{2} sin (2 x) d x

= - \frac{e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{10}

解答补常数

= - \frac{e ^{x} cos ( 2 x )}{5} + \frac{e ^{x} sin ( 2 x )}{10} + C

作图

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\int \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u

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\frac{d}{d x} (x^{2} lo g_{10} (x))