x3 | 1 | x4 | ||||
∫x2√R2−x2dx= | √R2−x2+ | ∫ | dx | |||
3 | 3 | √R2−x2 |
x3 | 1 | R2x2−x4−R2x2 | ||||
∫x2√R2−x2dx= | √R2−x2− | ∫ | dx | |||
3 | 3 | √R2−x2 |
x3 | 1 | R2 | x2 | ||||
√R2−x2− | ∫x2√R2−x2dx+ | ∫ | dx | ||||
3 | 3 | 3 | √R2−x2 |
4 | x3 | R2 | −x2+R2−R2 | ||||
∫x2√R2−x2dx= | √R2−x2− | ∫ | dx | ||||
3 | 3 | 3 | √R2−x2 |
4 | x3 | R2 | |||
∫x2√R2−x2dx= | √R2−x2− | ∫√R2−x2dx+ | |||
3 | 3 | 3 |
R4 | dx | ||
∫ | |||
3 | √R2−x2 |
x2 | ||
∫√R2−x2dx=x√R2−x2+∫ | dx | |
√R2−x2 |
R2−x2−R2 | ||
∫√R2−x2dx=x√R2−x2−∫ | dx | |
√R2−x2 |
dx | ||
∫√R2−x2dx=x√R2−x2−∫√R2−x2dx+R2∫ | ||
√R2−x2 |
dx | ||
2∫√R2−x2dx=x√R2−x2+R2∫ | dx | |
√R2−x2 |
1 | R2 | dx | ||||
∫√R2−x2dx= | x√R2−x2+ | ∫ | dx | |||
2 | 2 | √R2−x2 |
4 | x3 | ||
∫x2√R2−x2dx= | √R2−x2 | ||
3 | 3 |
R2 | 1 | R2 | dx | R4 | dx | |||||||
− | ( | x√R2−x2+ | ∫ | dx)+ | ∫ | |||||||
3 | 2 | 2 | √R2−x2 | 3 | √R2−x2 |
4 | 1 | R4 | dx | ||||
∫x2√R2−x2dx= | (2x3−R2x)√R2−x2+ | ∫ | dx | ||||
3 | 6 | 6 | √R2−x2 |
1 | R4 | x | ||||
∫x2√R2−x2dx= | (2x3−R2x)√R2−x2+ | arcsin( | )+C | |||
8 | 8 | R |
1 | ||
p= | ||
2 |
m+1 | 2+1 | 1 | |||
+p= | + | =2∊ℤ | |||
n | 2 | 2 |
R2−x2 | ||
więc stosujesz podstawienie t2= | ||
x2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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