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人気のある微分積分 >

integral of-11cos(ln(x))

解

\int - 11 cos (ln (x)) d x

解

- 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))) + C

解答ステップ

\int - 11 cos (ln (x)) d x

定数を除く:

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

= - 11 \cdot \int cos (ln (x)) d x

u置換積分法を適用する

= - 11 \cdot \int cos (u) e^{u} d u

部分積分を適用する

= - 11 (e^{u} sin (u) - \int e^{u} sin (u) d u)

部分積分を適用する

= - 11 (e^{u} sin (u) - (- e^{u} cos (u) - \int - e^{u} cos (u) d u))

定数を除く:

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

= - 11 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

このため

- 11 \cdot \int e^{u} cos (u) d u = - 11 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

分離する

\int e^{u} cos (u) d u

= - 11 (\frac{e ^{u} sin ( u )}{2} + \frac{e ^{u} cos ( u )}{2})

代用を戻す

u = ln (x)

= - 11 (\frac{e ^{l n (x)} sin ( ln ( x ) )}{2} + \frac{e ^{l n (x)} cos ( ln ( x ) )}{2})

簡素化

- 11 (\frac{e ^{l n (x)} sin ( ln ( x ) )}{2} + \frac{e ^{l n (x)} cos ( ln ( x ) )}{2}) : - 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))

= - 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))

定数を解答に追加する

= - 11 (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))) + C

グラフ

人気の例

tangent f(x)= x/(x+1),\at x=0

t an g e n t f (x) = \frac{x}{x + 1}, at x = 0

derivative f(x)=(2x-1)(6x-6)^{-1}

d er i v a t i v e f (x) = (2 x - 1) (6 x - 6)^{- 1}

derivative f(x)=(x^2)/(x+2)

d er i v a t i v e f (x) = \frac{x ^{2}}{x + 2}

(tan(y))dx-(cot(x))dy=0

(tan (y)) d x - (cot (x)) d y = 0

integral of (x-9)/((x+5)(x-2))

\int \frac{x - 9}{( x + 5 ) ( x - 2 )} d x