ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral of 4sin^2(xco)s^3x

解

\int 4 sin^{2} (x co) s^{3} x d x

解

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

解答ステップ

\int 4 sin^{2} (x co) s^{3} x d x

定数を除く:

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

= 4 s^{3} \cdot \int sin^{2} (co x) x d x

u置換積分法を適用する

= 4 s^{3} \cdot \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

定数を除く:

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

= 4 s^{3} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

部分積分を適用する

= 4 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u)

\int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))

= 4 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u)))

代用を戻す

u = co x

= 4 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x)))

簡素化

4 s^{3} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x))) : \frac{s ^{3}}{2 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

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

定数を解答に追加する

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

グラフ

人気の例

integral of sqrt(17)

\int 17 d x

integral from-6 to 0 of (2+sqrt(36-x^2))

\int_{- 6}^{0} (2 + 36 - x^{2}) d x

(\partial)/(\partial y)(x^2y^2z)

\frac{\partial}{\partial y} (x^{2} y^{2} z)

y^'+6y=t+e^{-5t}

y^{^{'}} + 6 y = t + e^{- 5 t}

integral of sin^4(x)*cos^4(x)

\int sin^{4} (x) \cdot cos^{4} (x) d x