integral from 1 to 2 of sin(2θ)cos(3θ)

解

\int_{1}^{2} sin (2 θ) cos (3 θ) d θ

\frac{1}{2} (\frac{1}{5} (- cos (10) + cos (5)) - cos (1) + cos (2))

十進法表記

- 0.36595 \dots

解答ステップ

\int_{1}^{2} sin (2 θ) cos (3 θ) d θ

次の恒等を使用する:

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

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

定数を除く:

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

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

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_{1}^{2} sin (2 θ + 3 θ) d θ + \int_{1}^{2} sin (2 θ - 3 θ) d θ)

\int_{1}^{2} sin (2 θ + 3 θ) d θ = \frac{1}{5} (- cos (10) + cos (5))

\int_{1}^{2} sin (2 θ - 3 θ) d θ = - cos (1) + cos (2)

= \frac{1}{2} (\frac{1}{5} (- cos (10) + cos (5)) - cos (1) + cos (2))