```python
from IPython.display import display, Latex
import sympy as sy

u, t, d, r = sy.symbols('u t d r')
sy.init_printing(use_unicode=True)

B = (u / 2) * (((r ** 2) * sy.sin(t))/((r ** 2 + d ** 2 - 2 * r * d * sy.cos(t))**(3/2)))

int1 = sy.Integral(B, t)
display(Latex(f"$${sy.latex(B)}={sy.latex(int1.doit())}={sy.latex(sy.integrate(B, t))}$$"))

```


$$\frac{r^{2} u \sin{\left(t \right)}}{2 \left(d^{2} - 2 d r \cos{\left(t \right)} + r^{2}\right)^{1.5}}=\frac{r^{2} u \left(\begin{cases} - \frac{\cos{\left(t \right)}}{\left(r^{2}\right)^{1.5}} & \text{for}\: d = 0 \\- \frac{\cos{\left(t \right)}}{\left(d^{2}\right)^{1.5}} & \text{for}\: r = 0 \\- \frac{1.0}{d r \left(d^{2} - 2 d r \cos{\left(t \right)} + r^{2}\right)^{0.5}} & \text{otherwise} \end{cases}\right)}{2}=\frac{r^{2} u \left(\begin{cases} - \frac{\cos{\left(t \right)}}{\left(r^{2}\right)^{1.5}} & \text{for}\: d = 0 \\- \frac{\cos{\left(t \right)}}{\left(d^{2}\right)^{1.5}} & \text{for}\: r = 0 \\- \frac{1.0}{d r \left(d^{2} - 2 d r \cos{\left(t \right)} + r^{2}\right)^{0.5}} & \text{otherwise} \end{cases}\right)}{2}$$


389b6be018133bc69ae743f3abae0b77d9e659ca
ipynb
Jupyter Notebook
PythonStudy/MathCalc.ipynb
flyinglandlord/BUAA-SAE-DataStructure-2021
7eb686465b18b125115e03b39e6d83e4ed7a30cb
MIT
2022-03-31T01:43:24.000Z
2022-03-31T02:05:14.000Z
PythonStudy/MathCalc.ipynb
flyinglandlord/BUAA-SAE-DataStructure-2021
7eb686465b18b125115e03b39e6d83e4ed7a30cb
MIT
PythonStudy/MathCalc.ipynb
flyinglandlord/BUAA-SAE-DataStructure-2021
7eb686465b18b125115e03b39e6d83e4ed7a30cb
MIT
Qwen/Qwen-72B
1. YES
2. YES