1. **State the problem:** We need to evaluate the integral $$\int \tan^2(4x) \sec^2(2x) \, dx$$. 2. **Recall formulas and identities:** - The derivative of $\tan(u)$ is $\sec^2(u) \frac{du}{dx}$. - We can use substitution if parts of the integral match the derivative of a function. - Also, recall the identity $\tan^2(\theta) = \sec^2(\theta) - 1$. 3. **Rewrite the integral using the identity:** $$\int \tan^2(4x) \sec^2(2x) \, dx = \int (\sec^2(4x) - 1) \sec^2(2x) \, dx$$ 4. **Notice the mismatch in arguments:** The $\sec^2$ terms have different arguments ($4x$ and $2x$), so direct substitution is complicated. 5. **Try substitution for $u = \tan(4x)$:** - Then $du = 4 \sec^2(4x) dx$ or $dx = \frac{du}{4 \sec^2(4x)}$. - But the integral has $\sec^2(2x)$, which does not simplify with this substitution. 6. **Try substitution for $v = \tan(2x)$:** - Then $dv = 2 \sec^2(2x) dx$ or $dx = \frac{dv}{2 \sec^2(2x)}$. - Rewrite the integral: $$\int \tan^2(4x) \sec^2(2x) dx = \int \tan^2(4x) \sec^2(2x) \cdot dx = \int \tan^2(4x) \sec^2(2x) \cdot \frac{dv}{2 \sec^2(2x)} = \frac{1}{2} \int \tan^2(4x) dv$$ 7. **Express $\tan^2(4x)$ in terms of $v$: since $v = \tan(2x)$, $4x = 2 \cdot 2x$, so $\tan(4x) = \tan(2 \cdot 2x)$, which can be expressed using the double angle formula for tangent:** $$\tan(2\alpha) = \frac{2 \tan(\alpha)}{1 - \tan^2(\alpha)}$$ So, $$\tan(4x) = \tan(2 \cdot 2x) = \frac{2 \tan(2x)}{1 - \tan^2(2x)} = \frac{2v}{1 - v^2}$$ Therefore, $$\tan^2(4x) = \left(\frac{2v}{1 - v^2}\right)^2 = \frac{4v^2}{(1 - v^2)^2}$$ 8. **Substitute back into the integral:** $$\frac{1}{2} \int \frac{4v^2}{(1 - v^2)^2} dv = 2 \int \frac{v^2}{(1 - v^2)^2} dv$$ 9. **Simplify the integral:** We need to evaluate $$I = \int \frac{v^2}{(1 - v^2)^2} dv$$ Rewrite numerator: $$v^2 = (1 - (1 - v^2)) = 1 - (1 - v^2)$$ So, $$I = \int \frac{1 - (1 - v^2)}{(1 - v^2)^2} dv = \int \frac{1}{(1 - v^2)^2} dv - \int \frac{1 - v^2}{(1 - v^2)^2} dv = \int \frac{1}{(1 - v^2)^2} dv - \int \frac{1}{1 - v^2} dv$$ 10. **Evaluate each integral separately:** - The second integral is $$\int \frac{1}{1 - v^2} dv = \frac{1}{2} \ln \left| \frac{1 + v}{1 - v} \right| + C$$ - The first integral is $$\int \frac{1}{(1 - v^2)^2} dv$$ Use partial fractions or standard integral formulas: Recall that $$\int \frac{1}{(1 - v)^2 (1 + v)^2} dv$$ Partial fraction decomposition gives: $$\frac{1}{(1 - v)^2 (1 + v)^2} = \frac{A}{1 - v} + \frac{B}{(1 - v)^2} + \frac{C}{1 + v} + \frac{D}{(1 + v)^2}$$ Solving for coefficients (omitted here for brevity) yields: $$\int \frac{1}{(1 - v^2)^2} dv = -\frac{1}{2(1 - v)} + \frac{1}{2(1 + v)} + \frac{1}{4} \ln \left| \frac{1 + v}{1 - v} \right| + C$$ 11. **Combine results:** $$I = \left(-\frac{1}{2(1 - v)} + \frac{1}{2(1 + v)} + \frac{1}{4} \ln \left| \frac{1 + v}{1 - v} \right| \right) - \frac{1}{2} \ln \left| \frac{1 + v}{1 - v} \right| + C = -\frac{1}{2(1 - v)} + \frac{1}{2(1 + v)} - \frac{1}{4} \ln \left| \frac{1 + v}{1 - v} \right| + C$$ 12. **Recall the factor 2 outside the integral:** $$2I = -\frac{1}{1 - v} + \frac{1}{1 + v} - \frac{1}{2} \ln \left| \frac{1 + v}{1 - v} \right| + C$$ 13. **Substitute back $v = \tan(2x)$:** $$\int \tan^2(4x) \sec^2(2x) dx = -\frac{1}{1 - \tan(2x)} + \frac{1}{1 + \tan(2x)} - \frac{1}{2} \ln \left| \frac{1 + \tan(2x)}{1 - \tan(2x)} \right| + C$$ **Final answer:** $$\boxed{\int \tan^2(4x) \sec^2(2x) dx = -\frac{1}{1 - \tan(2x)} + \frac{1}{1 + \tan(2x)} - \frac{1}{2} \ln \left| \frac{1 + \tan(2x)}{1 - \tan(2x)} \right| + C}$$