INTEGRATION IN CALCULUS

Question

Evaluate the definite integral. I do not know how to add the boundaries here so I am just going to write them.

Boundary is from 0 to 1/√2

 

∫ arcsinx/(√(1-x²)) dx

 

Thanks

 

Answer 1

Let u = Arcsinx   Then du = (1/√(1-x²))dx  

 

                             when x=0, u = Arcsin0 = 0

                             when x=1/√2, u = Arcsin(1/√2)=π/4

                                                       

So, we have ∫_{(from u=0 to u=π/4)}udu

 

                = (1/2)u²_{(from u=0 to u=π/4)}

 

                   = π²/32