1. **Problem Statement:** We want to find the integral of an inverse trigonometric function, for example, $\int \arcsin(x) \, dx$. 2. **Formula and Rules:** Integration by parts is often used for inverse trig functions. Recall the formula: $$\int u \, dv = uv - \int v \, du$$ Choose $u$ as the inverse trig function and $dv$ as the remaining part. 3. **Step-by-step Integration of $\int \arcsin(x) \, dx$:** - Let $u = \arcsin(x)$, so $du = \frac{1}{\sqrt{1-x^2}} \, dx$. - Let $dv = dx$, so $v = x$. 4. **Apply integration by parts:** $$\int \arcsin(x) \, dx = x \arcsin(x) - \int x \cdot \frac{1}{\sqrt{1-x^2}} \, dx$$ 5. **Simplify the remaining integral:** $$\int \frac{x}{\sqrt{1-x^2}} \, dx$$ Use substitution: let $w = 1 - x^2$, then $dw = -2x \, dx$ or $-\frac{1}{2} dw = x \, dx$. 6. **Rewrite the integral:** $$\int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{x}{\sqrt{w}} \, dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-\frac{1}{2}} \, dw$$ 7. **Integrate:** $$-\frac{1}{2} \int w^{-\frac{1}{2}} \, dw = -\frac{1}{2} \cdot 2 w^{\frac{1}{2}} + C = -\sqrt{w} + C = -\sqrt{1-x^2} + C$$ 8. **Final answer:** $$\int \arcsin(x) \, dx = x \arcsin(x) + \sqrt{1-x^2} + C$$ This process can be adapted for other inverse trig functions using integration by parts and appropriate substitutions.