integral from 0 to pi of cos^4(x)

解答

\int_{0}^{π} cos^{4} (x) d x

\frac{3 π}{8}

十进制

1.17809 \dots

求解步骤

\int_{0}^{π} cos^{4} (x) d x

化简

cos^{4} (x) : cos^{3} (x) cos (x)

= \int_{0}^{π} cos^{3} (x) cos (x) d x

使用分布积分法

= [cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x]_{0}^{π}

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

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

化简

[cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))]_{0}^{π} : [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{π}

= [\frac{1}{32} (32 cos^{3} (x) sin (x) + 3 (4 x - sin (4 x)))]_{0}^{π}

带入上下限计算后得:

\frac{3 π}{8}

= \frac{3 π}{8}

\frac{d}{d x} (1 - sin (x) cos (x))

\int t csc^{2} (\frac{3 t ^{2}}{2}) d t

n = 4 \sum \infty \frac{1}{n ^{2}}

t an g e n t f (x) = \frac{6}{5 x + 2}, (- 1, - 2)

\int (\frac{sin ( 4 x )}{4}) - (\frac{sin ^{3} ( x )}{4}) d x