1. Enunciado del problema: Calcular la integral $$\int t \arcsin(t) \, dt$$. 2. Para resolver esta integral, usaremos integración por partes. Recordemos la fórmula: $$\int u \, dv = uv - \int v \, du$$ 3. Elegimos: - $$u = \arcsin(t)$$, entonces $$du = \frac{1}{\sqrt{1-t^2}} dt$$ - $$dv = t \, dt$$, entonces $$v = \frac{t^2}{2}$$ 4. Aplicamos la fórmula: $$\int t \arcsin(t) \, dt = \frac{t^2}{2} \arcsin(t) - \int \frac{t^2}{2} \cdot \frac{1}{\sqrt{1-t^2}} dt$$ 5. Simplificamos la integral restante: $$\int \frac{t^2}{2 \sqrt{1-t^2}} dt = \frac{1}{2} \int \frac{t^2}{\sqrt{1-t^2}} dt$$ 6. Para resolver $$\int \frac{t^2}{\sqrt{1-t^2}} dt$$, usamos la sustitución $$t = \sin(\theta)$$, entonces $$dt = \cos(\theta) d\theta$$ y $$\sqrt{1-t^2} = \cos(\theta)$$. 7. Reescribimos la integral: $$\int \frac{\sin^2(\theta)}{\cos(\theta)} \cdot \cos(\theta) d\theta = \int \sin^2(\theta) d\theta$$ 8. Usamos la identidad trigonométrica: $$\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$$ 9. Entonces: $$\int \sin^2(\theta) d\theta = \int \frac{1 - \cos(2\theta)}{2} d\theta = \frac{1}{2} \int (1 - \cos(2\theta)) d\theta = \frac{1}{2} \left( \theta - \frac{\sin(2\theta)}{2} \right) + C$$ 10. Volvemos a la variable original: - $$\theta = \arcsin(t)$$ - $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2t \sqrt{1-t^2}$$ 11. Por lo tanto: $$\int \frac{t^2}{\sqrt{1-t^2}} dt = \frac{1}{2} \left( \arcsin(t) - \frac{2t \sqrt{1-t^2}}{2} \right) + C = \frac{1}{2} \arcsin(t) - \frac{t \sqrt{1-t^2}}{2} + C$$ 12. Finalmente, la integral original es: $$\int t \arcsin(t) dt = \frac{t^2}{2} \arcsin(t) - \frac{1}{2} \left( \frac{1}{2} \arcsin(t) - \frac{t \sqrt{1-t^2}}{2} \right) + C = \frac{t^2}{2} \arcsin(t) - \frac{1}{4} \arcsin(t) + \frac{t \sqrt{1-t^2}}{4} + C$$ 13. Simplificando: $$\int t \arcsin(t) dt = \left( \frac{t^2}{2} - \frac{1}{4} \right) \arcsin(t) + \frac{t \sqrt{1-t^2}}{4} + C$$