Как решить интеграл dx/(8-4sinx+7cosx)

Take the integral:
integral 1/(-4 sin(x)+7 cos(x)+8) dx
For the integrand 1/(-4 sin(x)+7 cos(x)+8), 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) (-(8 u)/(u^2+1)+(7 (1-u^2))/(u^2+1)+8)) du
Simplify the integrand 2/((u^2+1) (-(8 u)/(u^2+1)+(7 (1-u^2))/(u^2+1)+8)) to get 2/(u^2-8 u+15):
= integral 2/(u^2-8 u+15) du
Factor out constants:
= 2 integral 1/(u^2-8 u+15) du
For the integrand 1/(u^2-8 u+15), complete the square:
= 2 integral 1/((u-4)^2-1) du
For the integrand 1/((u-4)^2-1), substitute s = u-4 and ds = du:
= 2 integral 1/(s^2-1) ds
Factor -1 from the denominator:
= -2 integral 1/(1-s^2) ds
The integral of 1/(1-s^2) is tanh^(-1)(s):
= -2 tanh^(-1)(s)+constant
Substitute back for s = u-4:
= 2 tanh^(-1)(4-u)+constant
Substitute back for u = tan(x/2):
= 2 tanh^(-1)(4-tan(x/2))+constant
Which is equivalent for restricted x values to:
Answer: |
| = log(5 cos(x/2)-sin(x/2))-log(3 cos(x/2)-sin(x/2))+constant