ctgx | ||
∫ | = | |
sinx |
1 | ||
−dt= | dx | |
sinx |
ctgx |
| cosx | |||||||||
= | = | ||||||||||
sin2x | sin2x | sin3x |
cosx | 1 | |||
∫ | dx = ∫ | dt | ||
sin3x | t3 |
1 | 1 | |||
a ta calka z kolei | = t−3 ze standardowego wzoru ∫ xn dx = | xn+1 | ||
t3 | n+1 |
1 | 1 | |||
∫ctg x | dx = − | ctg2x bo (ctg x)' = −1/sin2x. | ||
sin2 x | 2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
Kliknij po więcej przykładów | |
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Twój nick | |