1 | ||
Wystarczy zamienic secans na | i tez mozna liczyc, ale latwiej mozna sie pomylic | |
cosx |
u2−1 | ||
t= | ||
2u |
2u2−(u2−1) | ||
u−t= | ||
2u |
u2+1 | ||
u−t= | ||
2u |
2u*2u−2(u2−1) | ||
dt= | du | |
4u2 |
u2+1 | ||
dt= | du | |
2u2 |
u2+1 | u2+1 | ||
−∫ | du | ||
2u | 2u2 |
(u2+1)2 | ||
−∫ | du | |
4u3 |
u4+2u2+1 | ||
−∫ | du | |
4u3 |
1 | 1 | 1 | ||||
− | (∫udu+∫ | du+2∫ | du) | |||
4 | u3 | u |
1 | u2 | 1 | ||||
− | ( | − | +2ln|u|)+C | |||
4 | 2 | 2u2 |
1 | u4−1 | |||
− | ( | +4ln|u|)+C | ||
8 | u2 |
1 | u4−1 | |||
− | ( | +ln|u|)+C | ||
2 | 4u2 |
1 | (u2−1)(u2+1) | |||
− | ( | +ln|u|)+C | ||
2 | 2u*2u |
1 | u2−1 | u2+1 | |||
− | ( | +ln|u|)+C | |||
2 | 2u | 2u |
1 | ||
− | (t√1+t2+ln|t+√1+t2|)+C | |
2 |
1 | ||
=− | (cos(x)√1+cos(x)2+ln|cos(x)+√1+cos(x)2|)+C | |
2 |
1 | dr | |||
t=tg r, 1+tg2r= | , dt = | |||
cos2r | cos2r |
1 | ds | |||
calka = ∫ | dr = ∫ | , s=sin r, | ||
cos3 r | (1−s2)2 |
1 | 1 | 1 | 1 | 1 | ||||||
= | ∫ | + | + | + | )ds | |||||
4 | (1−s)2 | (1+s)2 | 1−s | 1−s |
1 | 1 | 1 | t | |||||
= | ( | − | + ln(1+s) − ln(1−s)), s= | |||||
4 | 1−s | 1+s | √+t2 |
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 | |