integral from 0 to pi of sin(6x)cos(3x)

解答

\int_{0}^{π} sin (6 x) cos (3 x) d x

\frac{4}{9}

十进制

0.44444 \dots

求解步骤

\int_{0}^{π} sin (6 x) cos (3 x) d x

利用以下特性:

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

= \int_{0}^{π} \frac{sin ( 6 x + 3 x ) + sin ( 6 x - 3 x )}{2} 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_{0}^{π} sin (6 x + 3 x) + sin (6 x - 3 x) d x

化简

= \frac{1}{2} \cdot \int_{0}^{π} sin (9 x) + sin (3 x) d x

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_{0}^{π} sin (9 x) d x + \int_{0}^{π} sin (3 x) d x)

\int_{0}^{π} sin (9 x) d x = \frac{2}{9}

\int_{0}^{π} sin (3 x) d x = \frac{2}{3}

= \frac{1}{2} (\frac{2}{9} + \frac{2}{3})

化简

\frac{1}{2} (\frac{2}{9} + \frac{2}{3}) : \frac{4}{9}

= \frac{4}{9}

\int - 2 e^{x} (e^{x} - 1)^{3} d x

d er i v a t i v e 9 arctan (x - 1 + x^{2})

s l o p e (19, - 16), (- 7, - 15)

x \to 2 lim (x^{3})

d er i v a t i v e 7 x e^{x}