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受欢迎的微积分 >

integral of xsin(x^2)cos(3x^2)

解答

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

解答

\frac{1}{4} (- \frac{1}{4} cos (4 x^{2}) + \frac{1}{2} cos (2 x^{2})) + C

求解步骤

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

使用换元积分法

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

利用以下特性:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \frac{1}{2} \cdot \int \frac{sin ( u + 3 u ) + sin ( u - 3 u )}{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 \frac{1}{2} \cdot \int sin (u + 3 u) + sin (u - 3 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} \cdot \frac{1}{2} (\int sin (u + 3 u) d u + \int sin (u - 3 u) d u)

\int sin (u + 3 u) d u = - \frac{1}{4} cos (4 u)

\int sin (u - 3 u) d u = \frac{1}{2} cos (2 u)

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{4} cos (4 u) + \frac{1}{2} cos (2 u))

u = x^{2}

代回

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{4} cos (4 x^{2}) + \frac{1}{2} cos (2 x^{2}))

化简

\frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{4} cos (4 x^{2}) + \frac{1}{2} cos (2 x^{2})) : \frac{1}{4} (- \frac{1}{4} cos (4 x^{2}) + \frac{1}{2} cos (2 x^{2}))

= \frac{1}{4} (- \frac{1}{4} cos (4 x^{2}) + \frac{1}{2} cos (2 x^{2}))

解答补常数

= \frac{1}{4} (- \frac{1}{4} cos (4 x^{2}) + \frac{1}{2} cos (2 x^{2})) + C

作图

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