integral from 0 to 2pi of cos^7(x)

解

\int_{0}^{2 π} cos^{7} (x) d x

0

解答ステップ

\int_{0}^{2 π} cos^{7} (x) d x

簡素化

cos^{7} (x) : cos^{6} (x) cos (x)

= \int_{0}^{2 π} cos^{6} (x) cos (x) d x

部分積分を適用する

= [cos^{6} (x) sin (x) - \int - 6 cos^{5} (x) sin^{2} (x) d x]_{0}^{2 π}

\int - 6 cos^{5} (x) sin^{2} (x) d x = - 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7})

= [cos^{6} (x) sin (x) - (- 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7}))]_{0}^{2 π}

簡素化

[cos^{6} (x) sin (x) - (- 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7}))]_{0}^{2 π} : [cos^{6} (x) sin (x) + 6 (\frac{1}{3} sin^{3} (x) - \frac{2}{5} sin^{5} (x) + \frac{1}{7} sin^{7} (x))]_{0}^{2 π}

= [cos^{6} (x) sin (x) + 6 (\frac{1}{3} sin^{3} (x) - \frac{2}{5} sin^{5} (x) + \frac{1}{7} sin^{7} (x))]_{0}^{2 π}

限度を計算する:

0

= 0