1. **Stating the problem:** We want to integrate inverse trigonometric functions using algebraic manipulation. 2. **Formula and rules:** Recall that inverse trigonometric functions like $\arcsin x$, $\arccos x$, and $\arctan x$ have derivatives: $$\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx} \arctan x = \frac{1}{1+x^2}$$ 3. **Example:** Integrate $\int \arcsin x \, dx$ by algebraic manipulation. 4. **Step 1: Use integration by parts:** Let $u = \arcsin x$, $dv = dx$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$, and $v = x$. 5. **Step 2: Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\int \arcsin x \, dx = x \arcsin x - \int x \cdot \frac{1}{\sqrt{1-x^2}} dx$$ 6. **Step 3: Simplify the integral:** $$\int \frac{x}{\sqrt{1-x^2}} dx$$ Use substitution $t = 1 - x^2$, so $dt = -2x dx$ or $-\frac{1}{2} dt = x dx$. 7. **Step 4: Substitute and integrate:** $$\int \frac{x}{\sqrt{1-x^2}} dx = \int \frac{x}{\sqrt{t}} dx = \int \frac{-\frac{1}{2} dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-\frac{1}{2}} dt$$ 8. **Step 5: Integrate:** $$-\frac{1}{2} \int t^{-\frac{1}{2}} dt = -\frac{1}{2} \cdot 2 t^{\frac{1}{2}} + C = -\sqrt{t} + C = -\sqrt{1-x^2} + C$$ 9. **Step 6: Final answer:** $$\int \arcsin x \, dx = x \arcsin x + \sqrt{1-x^2} + C$$ This shows how algebraic manipulation and substitution help integrate inverse trig functions.