integral from 0 to 2npi of sin^6(x)

解

\int_{0}^{2 nπ} sin^{6} (x) d x

- \frac{1}{6} sin^{5} (2 πn) cos (2 πn) + \frac{5 ( - sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) ) )}{24}

解答ステップ

\int_{0}^{2 nπ} sin^{6} (x) d x

積分換算を適用する

= - [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{2 nπ} + \frac{5}{6} \cdot \int_{0}^{2 nπ} sin^{4} (x) d x

\int_{0}^{2 nπ} sin^{4} (x) d x = \frac{- sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) )}{4}

= - [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{2 nπ} + \frac{5}{6} \cdot \frac{- sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) )}{4}

簡素化

- [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{2 nπ} + \frac{5}{6} \cdot \frac{- sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) )}{4} : - [\frac{1}{6} sin^{5} (x) cos (x)]_{0}^{2 πn} + \frac{5 ( - sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) ) )}{24}

= - [\frac{1}{6} sin^{5} (x) cos (x)]_{0}^{2 πn} + \frac{5 ( - sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) ) )}{24}

限度を計算する:

\frac{1}{6} sin^{5} (2 πn) cos (2 πn)

= - \frac{1}{6} sin^{5} (2 πn) cos (2 πn) + \frac{5 ( - sin ^{3} ( 2 πn ) cos ( 2 πn ) + 3 ( πn - \frac{1}{4} sin ( 4 πn ) ) )}{24}