integral of sin(a+bx)

解答

\int sin (a + b x) d x

- \frac{1}{b} cos (a + b x) + C

求解步骤

\int sin (a + b x) d x

使用换元积分法

= \int \frac{sin ( u )}{b} d u

提出常数:

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

= \frac{1}{b} \cdot \int sin (u) d u

使用积分常见定则:

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

= \frac{1}{b} (- cos (u))

u = a + b x

代回

= \frac{1}{b} (- cos (a + b x))

化简

= - \frac{1}{b} cos (a + b x)

解答补常数

= - \frac{1}{b} cos (a + b x) + C

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