Question

rezolvati va rog urmatoarea integrala

Answer 1

$I = \int\ {1/sin^{6}x} \, dx= -\int\ {(1/sin^{2}x)^{2}*(-1)/sin^{2}x} \, dx$
Notam: ctg x=t => -1/sin²x dx=dt
Daca t=ctgx, avem: t=cos x/sin x.
t²=cos²x/sin²x=(1-sin²x)/sin²x=1/sin²x - 1
t²+1=1/sin²x
$I=- \int\ {(t^{2}+1)^{2}} \, dt=- \int\ {t^{4}+2t^{2}+1} \, dt=-t^{5}/5-2t^{3}/3-t+C$
Si revenind la noatia initiala avem:
$I=-(ctg^{5}x)/5-(2ctg^{3}x)/3-ctgx+C$