1. **Problem statement:** Find the Maclaurin series expansion for $\arcsin x$ up to the fourth term. 2. **Formula and rules:** The Maclaurin series for $\arcsin x$ is given by the infinite sum: $$\arcsin x = \sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}$$ This series includes only odd powers of $x$. 3. **Calculate terms:** - For $n=0$: $$\frac{(0)!}{4^0 (0!)^2 (1)} x^{1} = x$$ - For $n=1$: $$\frac{2!}{4^1 (1!)^2 (3)} x^{3} = \frac{2}{4 \cdot 1 \cdot 3} x^3 = \frac{1}{6} x^3$$ - For $n=2$: $$\frac{4!}{4^2 (2!)^2 (5)} x^{5} = \frac{24}{16 \cdot 4 \cdot 5} x^5 = \frac{24}{320} x^5 = \frac{3}{40} x^5$$ - For $n=3$: $$\frac{6!}{4^3 (3!)^2 (7)} x^{7} = \frac{720}{64 \cdot 36 \cdot 7} x^7 = \frac{720}{16128} x^7 = \frac{5}{112} x^7$$ 4. **Write the series up to the fourth term:** $$\arcsin x = x + \frac{1}{6} x^3 + \frac{3}{40} x^5 + \frac{5}{112} x^7 + \cdots$$ 5. **Explanation:** The Maclaurin series expansion expresses $\arcsin x$ as a power series centered at 0. Each term involves factorials and powers of $x$. We calculated the first four terms explicitly to approximate $\arcsin x$ near zero. **Final answer:** $$\arcsin x \approx x + \frac{1}{6} x^3 + \frac{3}{40} x^5 + \frac{5}{112} x^7$$