integral of sin(2x+y)

解答

\int sin (2 x + y) d x

- \frac{1}{2} cos (y + 2 x) + C

求解步骤

\int sin (2 x + y) d x

使用换元积分法

= \int sin (u) \frac{1}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int sin (u) d u

使用积分常见定则:

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

= \frac{1}{2} (- cos (u))

u = y + 2 x

代回

= \frac{1}{2} (- cos (y + 2 x))

化简

= - \frac{1}{2} cos (y + 2 x)

解答补常数

= - \frac{1}{2} cos (y + 2 x) + C

\int \frac{x x ^{2} + 4}{x ^{2} - 1} d x

d er i v a t i v e f (x) = \frac{x}{4}

\int (1 - x^{- 0.52}) d x

\frac{\partial}{\partial x} (5 ln (\frac{3 x y}{7 z}))

\int_{0}^{\frac{π}{4}} 9 sec^{2} (x) d x