Final answer:
To evaluate the integral ∫(dx)(arcsin(x))^2, we can use integration by parts. Let u = arcsin(x) and dv = (dx)(arcsin(x)), then differentiate u to get du = (dx)/sqrt(1-x^2) and integrate dv to get v = (1/2)(arcsin(x))^2 + x*sqrt(1-x^2)/2. Using the integration by parts formula...
Step-by-step explanation:
To evaluate the integral ∫(dx)(arcsin(x))^2, we can use integration by parts. Let u = arcsin(x) and dv = (dx)(arcsin(x)), then differentiate u to get du = (dx)/sqrt(1-x^2) and integrate dv to get v = (1/2)(arcsin(x))^2 + x*sqrt(1-x^2)/2. Using the integration by parts formula, we have:
∫(dx)(arcsin(x))^2 = u*v - ∫v*du
Substituting the values of u, v, and du, we have:
∫(dx)(arcsin(x))^2 = arcsin(x)[(1/2)(arcsin(x))^2 + x*sqrt(1-x^2)/2] - ∫[(1/2)(arcsin(x))^2 + x*sqrt(1-x^2)/2]*(dx)/sqrt(1-x^2)
Now, we can simplify the second integral and integrate it:
∫[(1/2)(arcsin(x))^2 + x*sqrt(1-x^2)/2]*(dx)/sqrt(1-x^2) = ∫[(1/2)(arcsin(x))^2/sqrt(1-x^2)]*(dx) + ∫[x*sqrt(1-x^2)/2]/sqrt(1-x^2)]*(dx)
The first integral can be solved using trigonometric substitution while the second integral simplifies to x/2. Substituting the values back in, we get:
∫(dx)(arcsin(x))^2 = arcsin(x)[(1/2)(arcsin(x))^2 + x*sqrt(1-x^2)/2] - ∫[(1/2)(arcsin(x))^2/sqrt(1-x^2)]*(dx) - ∫[x/2]*(dx)
Simplifying the expression further gives us the final result.