1. **State the problem:** We need to find the indefinite integral $$\int (1 - 2x^2)^5 \, dx$$. 2. **Formula and substitution:** Use substitution for integrals of the form $$\int (f(x))^n f'(x) \, dx$$. Let $$u = 1 - 2x^2$$. 3. **Find $$du$$:** Differentiate $$u$$ with respect to $$x$$: $$$du = -4x \, dx$$$ 4. **Rewrite the integral:** We want to express the integral in terms of $$u$$ and $$du$$. However, the integral does not have an $$x$$ term outside to match $$du$$ directly, so we use binomial expansion or another method. 5. **Binomial expansion:** Expand $$(1 - 2x^2)^5$$ using the binomial theorem: $$$(a - b)^5 = \sum_{k=0}^5 \binom{5}{k} a^{5-k} (-b)^k$$ Here, $$a=1$$ and $$b=2x^2$$. 6. **Calculate each term:** $$$\sum_{k=0}^5 \binom{5}{k} 1^{5-k} (-2x^2)^k = \sum_{k=0}^5 \binom{5}{k} (-2)^k x^{2k}$$$ 7. **Write out terms:** $$$= 1 - 10x^2 + 40x^4 - 80x^6 + 80x^8 - 32x^{10}$$$ 8. **Integrate term-by-term:** $$$\int (1 - 10x^2 + 40x^4 - 80x^6 + 80x^8 - 32x^{10}) \, dx = x - \frac{10x^3}{3} + 8x^5 - 10x^7 + \frac{80x^9}{9} - \frac{32x^{11}}{11} + C$$$ 9. **Final answer:** $$$\int (1 - 2x^2)^5 \, dx = x - \frac{10}{3}x^3 + 8x^5 - 10x^7 + \frac{80}{9}x^9 - \frac{32}{11}x^{11} + C$$$ This is the antiderivative of the given function.