We are tasked with evaluating the integral:

$I = \int_{0}^{\pi/4} \frac{\sec^2 x}{1 - 9\tan^2 x} \, dx$

Step-by-step solution:

1. Substitution: Let: $u = \tan x \quad \Rightarrow \quad du = \sec^2 x \, dx.$ The limits change as follows:

   • When $x = 0$, $u = \tan 0 = 0$,
   • When $x = \pi/4$, $u = \tan(\pi/4) = 1$.

   The integral becomes: $I = \int_{0}^{1} \frac{1}{1 - 9u^2} \, du.$

2. Simplify the integrand: The term $\frac{1}{1 - 9u^2}$ is a standard form of a partial fraction decomposition. Notice that: $1 - 9u^2 = (1 - 3u)(1 + 3u).$ Using partial fractions: $\frac{1}{1 - 9u^2} = \frac{A}{1 - 3u} + \frac{B}{1 + 3u}.$ Solving for $A$ and $B$, we get: $A = \frac{1}{6}, \quad B = \frac{-1}{6}.$

   Thus: $\frac{1}{1 - 9u^2} = \frac{1}{6(1 - 3u)} - \frac{1}{6(1 + 3u)}.$

3. Rewrite the integral: Substitute the partial fractions into the integral: $I = \int_{0}^{1} \left[\frac{1}{6(1 - 3u)} - \frac{1}{6(1 + 3u)}\right] \, du.$

4. Integrate each term: Using the standard formula $\int \frac{1}{a - bu} \, du = -\frac{1}{b} \ln |a - bu| + C$, we compute: $\int \frac{1}{1 - 3u} \, du = -\frac{1}{3} \ln |1 - 3u|, \quad \int \frac{1}{1 + 3u} \, du = \frac{1}{3} \ln |1 + 3u|.$

   Combine these results: $I = \frac{1}{6} \left[-\frac{1}{3} \ln |1 - 3u| + \frac{1}{3} \ln |1 + 3u|\right]_{0}^{1}.$

5. Simplify and apply limits: Factor out $\frac{1}{18}$: $I = \frac{1}{18} \left[\ln |1 + 3u| - \ln |1 - 3u|\right]_{0}^{1}.$ Using the logarithm property $\ln a - \ln b = \ln \frac{a}{b}$, this becomes: $I = \frac{1}{18} \ln \left[\frac{1 + 3u}{1 - 3u}\right]_{0}^{1}.$

   Now substitute the limits:

   • At $u = 1$: $\frac{1 + 3(1)}{1 - 3(1)} = \frac{4}{-2} = -2$,
   • At $u = 0$: $\frac{1 + 3(0)}{1 - 3(0)} = \frac{1}{1} = 1$.

   Therefore: $I = \frac{1}{18} \left[\ln(-2) - \ln(1)\right].$

   Since $\ln(-2)$ is not defined in the real domain, this integral diverges or is defined only in the complex domain.

Would you like to explore further about complex evaluations or have questions?

---

Related Questions:

1. How does substitution simplify trigonometric integrals?
2. What are partial fractions, and when are they useful in integration?
3. Can the domain restrictions of $\ln(x)$ affect the integral’s result?
4. What happens to this integral if $-9\tan^2 x$ were replaced with $+9\tan^2 x$?
5. How is this integral evaluated in the context of complex numbers?

Tip: Always check the domain of your integrand before starting computations to confirm the integral converges.