integral from 0 to pi/2 of sin(x)cos(2x)

解答

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

- \frac{1}{3}

十进制

- 0.33333 \dots

求解步骤

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

利用以下特性:

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

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

化简

sin (x + 2 x) + sin (x - 2 x) : sin (3 x) - sin (x)

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

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

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

= \frac{1}{2} (\frac{1}{3} - 1)

化简

\frac{1}{2} (\frac{1}{3} - 1) : - \frac{1}{3}

= - \frac{1}{3}

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