How to solve these two questions?

1.

Consider the integral 1 = f_dm 1 3:2 To derive a formula for this integral, we could use the circle trigonometric substitution 9: = sin(u) . =-a the integral can be easily evaluated as a function of u : cos(u) =/ , = l: a + c - 1 sin2(u) And nally as a function of the original variable cc . 1::na Note: the Maple syntax for cos1(3) is arccos (2:). Since Let's take the derivative of sinh (a) . Since sinh (sinh - (2)) = differentiating both sides with respect to x leads us to cosh (sinh - (x)) d sinh (ac) = + Now using the identity cosh (a) = 1 + sinh (x) , we can simplify cosh(sinh -(x)) = 1 + sinh(sinh -1 (x)) = So d dx (sinh -1 (2) = Of course, your table of standard integrals will remind you that 1 dx = sinh (x) + C 2 2 + 1 So the table of integrals can help you remember derivatives, too