zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 14sin^3(xco)s^2x

解答

\int 14 sin^{3} (x co) s^{2} x d x

解答

s^{2} \frac{14}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x)) + C

求解步骤

\int 14 sin^{3} (x co) s^{2} x d x

提出常数:

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

= 14 s^{2} \cdot \int sin^{3} (co x) x d x

使用换元积分法

= 14 s^{2} \cdot \int \frac{u sin ^{3} ( u )}{c ^{2} o ^{2}} d u

提出常数:

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

= 14 s^{2} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{3} (u) d u

使用分布积分法

= 14 s^{2} \frac{1}{c ^{2} o ^{2}} (u (- cos (u) + \frac{1}{3} cos^{3} (u)) - \int - cos (u) + \frac{1}{3} cos^{3} (u) d u)

\int - cos (u) + \frac{1}{3} cos^{3} (u) d u = - \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)

= 14 s^{2} \frac{1}{c ^{2} o ^{2}} (u (- cos (u) + \frac{1}{3} cos^{3} (u)) - (- \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)))

u = co x

代回

= 14 s^{2} \frac{1}{c ^{2} o ^{2}} (co x (- cos (co x) + \frac{1}{3} cos^{3} (co x)) - (- \frac{2}{3} sin (co x) - \frac{1}{9} sin^{3} (co x)))

化简

14 s^{2} \frac{1}{c ^{2} o ^{2}} (co x (- cos (co x) + \frac{1}{3} cos^{3} (co x)) - (- \frac{2}{3} sin (co x) - \frac{1}{9} sin^{3} (co x))) : s^{2} \frac{14}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x))

= s^{2} \frac{14}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x))

解答补常数

= s^{2} \frac{14}{c ^{2} o ^{2}} (- co x cos (co x) + \frac{1}{3} co x cos^{3} (co x) + \frac{2}{3} sin (co x) + \frac{1}{9} sin^{3} (co x)) + C

作图

流行的例子

integral of x^2-19

integral of 2xsec^2(5x)

integral of x^2-4x+5

limit as x approaches 0-of x/(ln(x))