How to solve these 3 question?

1.

Let's evaluate the definite integral 12 1 1:] da:. 6 x/m236 Note that we have to be a little careful here, as this is an improper in... 9 on account of the denominator becoming 0 at one of the end o . We use the circular trig substitution :1: = f (u) = 6*sec(u) o . As a; ranges from 6 to 12, the new variable u ranges from a, to b, where: . the new lower bound a is the value such that f(a) = 6. This is a = 0 o . the new upper bound b is the value such that f(b) = 12. This is b = -. The integral can thus be re-written with respect to u: b I=fg(u)du where the integrand g(u) = E]. g. Hence the integral can be evaluated as I: e o. Evaluate the definite integral using a suitable substitution. The answer is: o\\I| h... H 4> A 8 N) + lI \\.._./ _ U! da: