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受欢迎的微积分 >

integral of 2cos(ln(x))

解答

\int 2 cos (ln (x)) d x

解答

x sin (ln (x)) + x cos (ln (x)) + C

求解步骤

\int 2 cos (ln (x)) d x

提出常数:

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

= 2 \cdot \int cos (ln (x)) d x

使用换元积分法

= 2 \cdot \int cos (u) e^{u} d u

使用分布积分法

= 2 (e^{u} sin (u) - \int e^{u} sin (u) d u)

使用分布积分法

= 2 (e^{u} sin (u) - (- e^{u} cos (u) - \int - e^{u} cos (u) d u))

提出常数:

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

= 2 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

因此

2 \cdot \int e^{u} cos (u) d u = 2 (e^{u} sin (u) - (- e^{u} cos (u) - (- \int e^{u} cos (u) d u)))

整理

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

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

u = ln (x)

代回

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

化简

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

= x sin (ln (x)) + x cos (ln (x))

解答补常数

= x sin (ln (x)) + x cos (ln (x)) + C

作图

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