AP Calculus BC - Integration

Question

Find the derivatives of each function:

1. h(x) = ₂∫^1/x arctan(t) dt
2. H(x) = ₂∫^{x^2} (w-1)/(√w) dw
3. g(x) = _{tan x} ∫^{x^3} √t sin (t) dt

Answer 1

h(x) = ₂∫^1/x arctan(t) dt

Start by integrating by parts

∫fg' = fg - ∫f'g

Now: f = arctan(t), and g = t

f'= 1/(t^2 +1), and g' = 1

= tarctan(t) - ∫t/(t^2 +1) dt

Solve the integral by u-substitution: u = t^2+1, du/dt = 2t,

1/2tdu = dt

= tarctan(t) - ∫1/2u du

= tarctan(t) - 1/2ln(u)

= tarctan(t) - 1/2ln(t^2+1); since u = t^2+1

boundaries: [tarctan(t) - 1/2ln(t^2 +1)] 1/x 2

= 1/xarctan(1/x) - 1/2ln(1/x^2 +1) - 2arctan(2) + 1/2ln(5) Answer

H(x) = ₂∫^{x^2} (w-1)/(√w) dw

∫ (√w - 1/√w) dw

Separating:

H(x) = ∫√w dw - ∫1/√w dw

= 2w^(3/2)/3 - 2w^(1/2)

= [ 2/3w^(3/2) - 2w^(1/2) ] x^2 2

= [ 2/3x^3 - 2x ] - [ 2/3(2)^3/2 - 2(2)^1/2]

= 2/3x^3 - 2x - 2/3(2)^3/2 + 2√2 Answer