Solve the integral 1 on top and zero on the bottom.

Question

¹₀∫sec√x/√x dx

Answer 1

Use u-substitution.

Let u=√x.

du = 1/2√xdx

dx = 2√xdu

Translate the integration limits from x-space to u-space.

Lower limit for u is √0 = 0

Upper limit for u is √1 = 1

So, translating the integral from x-space to u-space:

∫_o¹ sec(√x)/√x dx -> ∫_o¹ sec(u)/√x (2√xdu)

= 2 ∫_o¹ sec(u)du

= 2 ln(sec(u) + tan(u))₀¹

= 2 (ln(sec(1)+tan(1))-ln(sec(0)+tan(0)))

≈ 2 (ln(1.851+1.557)-ln(1+0)))

≈ 2 (ln(3.408)+ln(1))

≈ 2 (1.226 + 0)

≈ 2.452