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人気のある微分積分 >

integral from-pi to 0 of sin(3x)cos(2x)

解

\int_{- π}^{0} sin (3 x) cos (2 x) d x

解

- \frac{6}{5}

+1

十進法表記

- 1.2

解答ステップ

\int_{- π}^{0} sin (3 x) cos (2 x) d x

次の恒等を使用する:

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

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

定数を除く:

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

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{- π}^{0} sin (3 x + 2 x) d x = - \frac{2}{5}

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

= \frac{1}{2} (- \frac{2}{5} - 2)

簡素化

\frac{1}{2} (- \frac{2}{5} - 2) : - \frac{6}{5}

= - \frac{6}{5}

グラフ

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