1. **Problem statement:** Evaluate the integral $$\int \frac{dx}{5 + f\left(\frac{\pi}{2} - x\right)}$$ where $f(x) = \tan x$. 2. **Substitute the function:** Since $f(x) = \tan x$, we have $$f\left(\frac{\pi}{2} - x\right) = \tan\left(\frac{\pi}{2} - x\right).$$ 3. **Use the trigonometric identity:** $$\tan\left(\frac{\pi}{2} - x\right) = \cot x = \frac{1}{\tan x}.$$ 4. **Rewrite the integral:** $$\int \frac{dx}{5 + \cot x} = \int \frac{dx}{5 + \frac{1}{\tan x}} = \int \frac{dx}{5 + \frac{1}{\tan x}}.$$ 5. **Combine the denominator:** $$5 + \frac{1}{\tan x} = \frac{5 \tan x + 1}{\tan x}.$$ 6. **Rewrite the integral:** $$\int \frac{dx}{\frac{5 \tan x + 1}{\tan x}} = \int \frac{\tan x}{5 \tan x + 1} dx.$$ 7. **Let $t = \tan x$, then $dt = \sec^2 x dx = (1 + \tan^2 x) dx = (1 + t^2) dx$, so:** $$dx = \frac{dt}{1 + t^2}.$$ 8. **Substitute into the integral:** $$\int \frac{t}{5t + 1} \cdot \frac{dt}{1 + t^2} = \int \frac{t}{(5t + 1)(1 + t^2)} dt.$$ 9. **Use partial fraction decomposition:** Set $$\frac{t}{(5t + 1)(1 + t^2)} = \frac{A}{5t + 1} + \frac{Bt + C}{1 + t^2}.$$ 10. **Multiply both sides by $(5t + 1)(1 + t^2)$:** $$t = A(1 + t^2) + (Bt + C)(5t + 1).$$ 11. **Expand the right side:** $$t = A + A t^2 + 5 B t^2 + B t + 5 C t + C.$$ 12. **Group terms by powers of $t$:** $$t = (A + C) + (B + 5 C) t + (A + 5 B) t^2.$$ 13. **Equate coefficients:** - Constant term: $A + C = 0$ - Coefficient of $t$: $B + 5 C = 1$ - Coefficient of $t^2$: $A + 5 B = 0$ 14. **Solve the system:** From $A + C = 0$, we get $C = -A$. From $A + 5 B = 0$, we get $A = -5 B$. Substitute into $B + 5 C = 1$: $$B + 5(-A) = 1 \Rightarrow B - 5 A = 1.$$ Replace $A$ with $-5 B$: $$B - 5(-5 B) = B + 25 B = 26 B = 1 \Rightarrow B = \frac{1}{26}.$$ Then $A = -5 B = -\frac{5}{26}$ and $C = -A = \frac{5}{26}$. 15. **Rewrite the integral:** $$\int \left( \frac{-\frac{5}{26}}{5 t + 1} + \frac{\frac{1}{26} t + \frac{5}{26}}{1 + t^2} \right) dt = -\frac{5}{26} \int \frac{dt}{5 t + 1} + \frac{1}{26} \int \frac{t}{1 + t^2} dt + \frac{5}{26} \int \frac{dt}{1 + t^2}.$$ 16. **Integrate each term:** - $$\int \frac{dt}{5 t + 1} = \frac{1}{5} \ln|5 t + 1| + C.$$ - $$\int \frac{t}{1 + t^2} dt = \frac{1}{2} \ln(1 + t^2) + C.$$ - $$\int \frac{dt}{1 + t^2} = \arctan t + C.$$ 17. **Substitute back:** $$= -\frac{5}{26} \cdot \frac{1}{5} \ln|5 t + 1| + \frac{1}{26} \cdot \frac{1}{2} \ln(1 + t^2) + \frac{5}{26} \arctan t + C$$ $$= -\frac{1}{26} \ln|5 \tan x + 1| + \frac{1}{52} \ln(1 + \tan^2 x) + \frac{5}{26} \arctan(\tan x) + C.$$ 18. **Simplify using $1 + \tan^2 x = \sec^2 x$ and $\arctan(\tan x) = x$ (for principal values):** $$= -\frac{1}{26} \ln|5 \tan x + 1| + \frac{1}{52} \ln(\sec^2 x) + \frac{5}{26} x + C.$$ 19. **Final answer:** $$\boxed{\int \frac{dx}{5 + \tan\left(\frac{\pi}{2} - x\right)} = -\frac{1}{26} \ln|5 \tan x + 1| + \frac{1}{52} \ln(\sec^2 x) + \frac{5}{26} x + C}.$$