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人気のある微分積分 >

integral of sin^2(3xco)s^23x

解

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

解

\frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x)) + C

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

= 3 s^{2} \cdot \int \frac{u sin ^{2} ( u )}{9 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

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

部分積分を適用する

= 3 s^{2} \frac{1}{9 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))

= 3 s^{2} \frac{1}{9 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 = 3 co x

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} \cdot 3 co x (3 co x - \frac{1}{2} sin (2 \cdot 3 co x)) - \frac{1}{2} (\frac{( 3 co x ) ^{2}}{2} + \frac{1}{4} cos (2 \cdot 3 co x)))

簡素化

3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} \cdot 3 co x (3 co x - \frac{1}{2} sin (2 \cdot 3 co x)) - \frac{1}{2} (\frac{( 3 co x ) ^{2}}{2} + \frac{1}{4} cos (2 \cdot 3 co x))) : \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x))

= \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x))

定数を解答に追加する

= \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x)) + C

グラフ

人気の例

(dy)/(dx)=e^{y+x}

\frac{d y}{d x} = e^{y + x}

d/(dv)(usin(v))

\frac{d}{d v} (u sin (v))

(\partial)/(\partial y)(ln(5+x^2y^2))

\frac{\partial}{\partial y} (ln (5 + x^{2} y^{2}))

tangent f(x)=4x-x^2,\at x=2

t an g e n t f (x) = 4 x - x^{2}, at x = 2

勾配 s=t^3-t^2

s l o p e s = t^{3} - t^{2}