integral from 0 to 1/2 of sin^4(x)

解

\int_{0}^{\frac{1}{2}} sin^{4} (x) d x

- \frac{1}{4} sin^{3} (\frac{1}{2}) cos (\frac{1}{2}) + \frac{3}{8} (\frac{1}{2} - \frac{1}{2} sin (1))

十進法表記

0.00554 \dots

解答ステップ

\int_{0}^{\frac{1}{2}} sin^{4} (x) d x

積分換算を適用する

= - [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{1}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{1}{2}} sin^{2} (x) d x

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

= - [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{1}{2}} + \frac{3}{4} \cdot \frac{1}{2} (\frac{1}{2} - \frac{1}{2} sin (1))

簡素化

- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{1}{2}} + \frac{3}{4} \cdot \frac{1}{2} (\frac{1}{2} - \frac{1}{2} sin (1)) : - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{1}{2}} + \frac{3}{8} (\frac{1}{2} - \frac{1}{2} sin (1))

= - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{1}{2}} + \frac{3}{8} (\frac{1}{2} - \frac{1}{2} sin (1))

限度を計算する:

\frac{1}{4} sin^{3} (\frac{1}{2}) cos (\frac{1}{2})

= - \frac{1}{4} sin^{3} (\frac{1}{2}) cos (\frac{1}{2}) + \frac{3}{8} (\frac{1}{2} - \frac{1}{2} sin (1))