integral of q^2cos(4q)

解答

\int q^{2} cos (4 q) d q

\frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q)) + C

求解步骤

\int q^{2} cos (4 q) d q

使用换元积分法

= \int \frac{u ^{2} cos ( u )}{64} d u

提出常数:

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

= \frac{1}{64} \cdot \int u^{2} cos (u) d u

使用分布积分法

= \frac{1}{64} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{64} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

u = 4 q

代回

= \frac{1}{64} ((4 q)^{2} sin (4 q) - 2 (- 4 q cos (4 q) + sin (4 q)))

化简

\frac{1}{64} ((4 q)^{2} sin (4 q) - 2 (- 4 q cos (4 q) + sin (4 q))) : \frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q))

= \frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q))

解答补常数

= \frac{1}{32} (8 q^{2} sin (4 q) - sin (4 q) + 4 q cos (4 q)) + C

t an g e n t 3 x^{2} + 3 x - 5

\frac{\partial}{\partial x} (\frac{2 x}{y})

d er i v a t i v e y = x^{4} e^{3 x}

n = 5 \sum \infty \frac{1}{n ^{2} + n}

\int_{0}^{\frac{π}{3}} cos (4 x) d x