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受欢迎的微积分 >

integral of-16sin(4x-pi/2)

解答

\int - 16 sin (4 x - \frac{π}{2}) d x

解答

4 sin (4 x) + C

求解步骤

\int - 16 sin (4 x - \frac{π}{2}) d x

提出常数:

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

= - 16 \cdot \int sin (4 x - \frac{π}{2}) d x

使用三角恒等式改写

= - 16 \cdot \int - cos (4 x) d x

提出常数:

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

= - 16 (- \int cos (4 x) d x)

使用换元积分法

= - 16 (- \int cos (u) \frac{1}{4} d u)

提出常数:

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

= - 16 (- \frac{1}{4} \cdot \int cos (u) d u)

使用积分常见定则:

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

= - 16 (- \frac{1}{4} sin (u))

u = 4 x

代回

= - 16 (- \frac{1}{4} sin (4 x))

化简

- 16 (- \frac{1}{4} sin (4 x)) : 4 sin (4 x)

= 4 sin (4 x)

解答补常数

= 4 sin (4 x) + C

作图

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