1. **Stating the problem:** We want to find the integral $$\int \frac{1}{1+\cos^4 t} \, dt$$ and analyze its behavior for graphing. 2. **Formula and approach:** There is no simple elementary antiderivative for $$\frac{1}{1+\cos^4 t}$$, so we consider rewriting or numerical methods for graphing. 3. **Rewrite the integrand:** Note that $$\cos^4 t = (\cos^2 t)^2$$ and use the identity $$\cos^2 t = \frac{1+\cos 2t}{2}$$. 4. Substitute: $$\cos^4 t = \left(\frac{1+\cos 2t}{2}\right)^2 = \frac{1 + 2\cos 2t + \cos^2 2t}{4}$$ 5. Use $$\cos^2 2t = \frac{1+\cos 4t}{2}$$: $$\cos^4 t = \frac{1 + 2\cos 2t + \frac{1+\cos 4t}{2}}{4} = \frac{2 + 4\cos 2t + 1 + \cos 4t}{8} = \frac{3 + 4\cos 2t + \cos 4t}{8}$$ 6. So the integrand becomes: $$\frac{1}{1 + \cos^4 t} = \frac{1}{1 + \frac{3 + 4\cos 2t + \cos 4t}{8}} = \frac{1}{\frac{8 + 3 + 4\cos 2t + \cos 4t}{8}} = \frac{8}{11 + 4\cos 2t + \cos 4t}$$ 7. The integral is now: $$\int \frac{8}{11 + 4\cos 2t + \cos 4t} \, dt$$ 8. This integral does not simplify easily to elementary functions, so for graphing, numerical integration or plotting the integrand $$y = \frac{1}{1+\cos^4 t}$$ is recommended. 9. **Graph features:** The function is periodic with period $$\pi$$ because $$\cos^4 t$$ has period $$\pi$$. 10. The function is always positive since denominator $$1 + \cos^4 t \geq 1$$. 11. The minimum value of the denominator is 1 (when $$\cos t = 0$$), so maximum of integrand is 1. 12. The maximum value of denominator is $$1 + 1 = 2$$ (when $$\cos t = \pm 1$$), so minimum of integrand is $$\frac{1}{2} = 0.5$$. 13. The graph oscillates between 0.5 and 1, positive, periodic. 14. For plotting, use $$y = \frac{1}{1+\cos^4 t}$$ with domain $$t \in [-2\pi, 2\pi]$$ to see multiple periods. **Final answer:** The integral does not have a simple closed form in elementary functions, but the integrand simplifies to $$y = \frac{8}{11 + 4\cos 2t + \cos 4t}$$ which is positive and periodic with period $$\pi$$.