definite integrals

I = ∫1/[x^2 √(36-x^2 )] dxLet x = 6 sin u=> dx = 6 cos u duwhen x = 3√3 => sin u = √3/2 => u = π/3when x = 3√2=> sin u=√2/2 => u = π/4I = ∫(6 cos u)/[36 sin^2 u (6 cos u)] du = 1/36 ∫csc^2 u du= [-1/36 cot u ] with lower limit = π/4 and upper limit = π/3= 1/36 (1-1/√3) I = ∫1/√[(x+1)^2+16] dxLet x+1 = 4 tan udx = 4 sec^2 u duwhen x = 3, tan u = 1 => u = π/4when x=-1, tan u = 0 => u = 0I = ∫ 4 sec^2 u/(4 sec u) du = ∫sec u du = [ln |sec u + tan u| ] with lower limit = 0 and upper limit = π/4= ln (1+√2)