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Un pasado de gloria, un presente de luz. <strong>Revista</strong> IC <strong>6°</strong>A 2018<br />
2.-<br />
du<br />
u 2 ± a 2 = ln (u + u2 ± a 2 ) + c<br />
=<br />
1<br />
u 2 ± a du =<br />
2<br />
a sec u tan u<br />
a 2 sec 2 u − 1 du<br />
= a u 2 sec u tan u<br />
sec 2 u − 1 du = a u 2 sec u tan u<br />
−1 + 1 + tan 2 u du<br />
= a u 2 sec u tan u<br />
tan y<br />
= a 1<br />
ln |tan [arc sec<br />
u2 = ln |<br />
u 2 − a 2 + u<br />
| = ln |<br />
a<br />
du = a ln|tan u + sec u |<br />
u2 a u ]| + sec [arc sec 1 a u ]<br />
u 2 − a 2 + u<br />
| + c<br />
a<br />
3.-<br />
du<br />
a 2 − u = arc sen u 2 a + c<br />
1<br />
=<br />
a 2 − u 2 du<br />
a cos u<br />
=<br />
u 2 −sen 2 u + 1 = a cos u<br />
u 2 −sen 2 u + 1<br />
= a cos u<br />
u 2 cos 2 u du = a cos u<br />
u 2 cos u du<br />
= a u 2 1 du = a u 2 1 u = a u 2 1 arc sen 1 a u = arc sen (u a )<br />
= arc sen ( u a ) + c<br />
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