x^4=u
4x^3dx=du
x^3dx=du/4
∫ x^3dx/(1+x^8)= ∫ x^3/(1+(x^4)^2)= ∫ (du/4)/(1+u^2)=(1/4)arctgu
поэтому:
∫ ^(1)_(0)x^3dx/(1+x^8)=(1/4)∫ ^(1)_(0)d(x^4)/(1+(x^4)^2)=
=(1/4)(arctgx^4|^(1)_(0)=(1/4)arctg1=(1/4)*(π/4)= [b]π/16[/b]
4.
∫ ^(π/6)_(0)e^(sinx)cosxdx=∫ ^(π/6)_(0)e^(sinx) d(sinx)=
=e^(sinx)|^(π/6)_(0)=e^(sin(π/6))-e^(sin0)=e^(1/2)-e^(0)= [b]sqrt(e)-1[/b]