integral of t^3cos(t^2)

解答

\int t^{3} cos (t^{2}) d t

\frac{1}{2} (t^{2} sin (t^{2}) + cos (t^{2})) + C

求解步骤

\int t^{3} cos (t^{2}) d t

使用换元积分法

= \int \frac{u cos ( u )}{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 u cos (u) d u

使用分布积分法

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

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

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

u = t^{2}

代回

= \frac{1}{2} (t^{2} sin (t^{2}) - (- cos (t^{2})))

化简

= \frac{1}{2} (t^{2} sin (t^{2}) + cos (t^{2}))

解答补常数

= \frac{1}{2} (t^{2} sin (t^{2}) + cos (t^{2})) + C

x \to 5 + lim (\frac{5 - x}{∣ 5 - x ∣})

d er i v a t i v e ln (4 x + 7)

\int \frac{( 3 + e ^{x} ) ^{2}}{e ^{x}} d x

\frac{\partial}{\partial x} (- y sin (x))

\frac{d}{d x} (\frac{sin ( 2 x )}{tan ( x )})