Примеры по высшей математике.Нужна помощь.

Possible intermediate steps:
integral 1/(5+5 cos(x)+3 sin(x)) dx
For the integrand 1/(3 sin(x)+5 cos(x)+5), substitute u = tan(x/2) and du = 1/2 sec^2(x/2) dx. Then transform the integrand using the substitutions sin(x) = (2 u)/(u^2+1), cos(x) = (1-u^2)/(u^2+1) and dx = (2 du)/(u^2+1):
= integral 2/((u^2+1) ((6 u)/(u^2+1)+(5 (1-u^2))/(u^2+1)+5)) du
Simplify the integrand 2/((u^2+1) ((6 u)/(u^2+1)+(5 (1-u^2))/(u^2+1)+5)) to get 1/(3 u+5):
= integral 1/(3 u+5) du
For the integrand 1/(3 u+5), substitute s = 3 u+5 and ds = 3 du:
= 1/3 integral 1/s ds
The integral of 1/s is log(s):
= (log(s))/3+constant
Substitute back for s = 3 u+5:
= 1/3 log(3 u+5)+constant
Substitute back for u = tan(x/2):
= 1/3 log(3 tan(x/2)+5)+constant
Which is equivalent for restricted x values to:
= 1/3 (log(3 sin(x/2)+5 cos(x/2))-log(cos(x/2)))+constant

Possible intermediate steps:
integral 1/(1+cos(x)+2 sin(x)) dx
For the integrand 1/(2 sin(x)+cos(x)+1), substitute u = tan(x/2) and du = 1/2 sec^2(x/2) dx. Then transform the integrand using the substitutions sin(x) = (2 u)/(u^2+1), cos(x) = (1-u^2)/(u^2+1) and dx = (2 du)/(u^2+1):
= integral 2/((u^2+1) ((4 u)/(u^2+1)+(1-u^2)/(u^2+1)+1)) du
Simplify the integrand 2/((u^2+1) ((4 u)/(u^2+1)+(1-u^2)/(u^2+1)+1)) to get 1/(2 u+1):
= integral 1/(2 u+1) du
For the integrand 1/(2 u+1), substitute s = 2 u+1 and ds = 2 du:
= 1/2 integral 1/s ds
The integral of 1/s is log(s):
= (log(s))/2+constant
Substitute back for s = 2 u+1:
= 1/2 log(2 u+1)+constant
Substitute back for u = tan(x/2):
= 1/2 log(2 tan(x/2)+1)+constant
Which is equivalent for restricted x values to:
= 1/2 (log(2 sin(x/2)+cos(x/2))-log(cos(x/2)))+constant