# 微積分の計算について　　　N0.3　不定積分の内容－2  置換積分

###  学籍番号[＿＿＿＿＿＿＿＿＿]クラス[＿＿＿＿＿]　クラス番号[＿＿＿＿＿] 名前[＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿]

##### 積分の手法　（１）置換積分法

$$  t= g(x) または　x=u(t) $$

$$ \int f(x)dx = \int f(u)\frac{du}{dt}dt $$


```python
from sympy import *
x, n , y, a = symbols('x n y a')
init_printing()
m ='3-2'
i =0
```

 ### 例題ー１


```python
expr = (x+1)**2
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python
expr = sqrt(x+1)
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python
expr = cos(2*x)
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
expr = cos(x)**3
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python
itg.doit()
```


```python
expr = x/(x+1)
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python
expr = 1/(x-1)**2
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python
expr = x/sqrt(x**2+a**2)
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python
expr = sin(x)**2 * cos(x)
itg = Integral(expr,x)
i=i+1
print( 'No.',m,'---',i)
itg
```


```python
itg.doit()
```


```python

```


```python

```

27d58dbc7b35fc2e9a76cadc2db7194e6d8197dd
ipynb
Jupyter Notebook
04_20181025-sekibun-3-1-Ex&ans.ipynb
kt-pro-git-1/Calculus_Differential_Equation-public
d5deaf117e6841c4f6ceb53bc80b020220fd4814
MIT
2019-07-10T11:33:18.000Z
2019-07-10T11:33:18.000Z
04_20181025-sekibun-3-1-Ex&ans.ipynb
kt-pro-git-1/Calculus_Differential_Equation-public
d5deaf117e6841c4f6ceb53bc80b020220fd4814
MIT
04_20181025-sekibun-3-1-Ex&ans.ipynb
kt-pro-git-1/Calculus_Differential_Equation-public
d5deaf117e6841c4f6ceb53bc80b020220fd4814
MIT
Qwen/Qwen-72B
1. YES
2. YES