∫ from 0 - pi/4 of the function sec theta tan theta d theta, evalutate the integral? I NEED HELP?

Question

This is an integral with 0 on the bottom of it and pi/4 on top just to make sure you understand... and it needs to be evalutated

Answer 1

∫(0 to π/4) secθtanθdθ = ∫(0 to π/4) sinθ/(cosθ^2)dθ
Let u = cosθ
So du = sinθdθ
When θ = 0, u = 1
When θ = π/4, u = 1/√2

So ∫(0 to π/4) secθtanθdθ
= ∫(1 to 1/√2) 1/u^2 du
= -u^-1 (from 1 to 1/√2)
= -1^-1 - -(1/√2)^-1
= √2 - 1

Answer 2

∫secө tanө dө
= secө, from 0 to pi/4
= √2 - 1

Answer 3

enable tan theta = a sec^2theta d theta = da quintessential sec^2 theta / tan theta = quintessential da/a quintessential da/a = log (a) + c for this reason quintessential sec^2 theta /tan theta = log tan theta.+c

Answer 4

remember sec t = 1/cos t so you have the Int sint/cos^2t dt
cos t=z so -sin t dt =dz and the int becomes
-Int 1/z^2 dz = 1/z = 1/cos t( 0,pi/4) = sqrt(2)-1

Answer 5

∫(0..π/4) sec t tan t dt = sec t(0..π/4) = sec π/4 - 1
= √2-1.