integral of 4xcos(9x)

解答

\int 4 x cos (9 x) d x

\frac{4}{81} (9 x sin (9 x) + cos (9 x)) + C

求解步骤

\int 4 x cos (9 x) d x

提出常数:

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

= 4 \cdot \int x cos (9 x) d x

使用换元积分法

: \int \frac{u cos ( u )}{81} d u

= 4 \cdot \int \frac{u cos ( u )}{81} d u

提出常数:

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

= 4 \cdot \frac{1}{81} \cdot \int u cos (u) d u

使用分布积分法

: u sin (u) - \int sin (u) d u

= 4 \cdot \frac{1}{81} (u sin (u) - \int sin (u) d u)

\int sin (u) d u = - cos (u)

= 4 \cdot \frac{1}{81} (u sin (u) - (- cos (u)))

u = 9 x

代回

= 4 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x)))

化简

4 \cdot \frac{1}{81} (9 x sin (9 x) - (- cos (9 x))) : \frac{4}{81} (9 x sin (9 x) + cos (9 x))

= \frac{4}{81} (9 x sin (9 x) + cos (9 x))

解答补常数

= \frac{4}{81} (9 x sin (9 x) + cos (9 x)) + C

\frac{d}{d x} (\frac{1}{8} ln (\frac{3 x - 4}{3 x + 4}))

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