Question

evaluate the integral:

integral from -pi/4 to 0 for the function 6sec^3x dx.

it has to be an exact answer and i did it and keep getting it wrong. I got
4sqrt(2)-4ln(-sqrt(2)+1)

Answer 1

According to the Wolfram integrator , this looks like a messy integration
http://integrals.wolfram.com/index.jsp?expr=6%2F%28cos%28x%29%29%5E3&random=false

looks like multiple integration by parts, arhhh!

Answer 2

u = sec x,
dv = sec^2 x dx

du = secx tanx dx
v = tan x

∫sec^3 x dx
= ∫u dv = uv - ∫v du
= secx tanx - ∫secx tan^2 x dx
= secx tanx - ∫(secx (sec^2 x - 1) dx
= secx tanx - ∫sec^3 x dx + ∫secx dx
so,
2∫sec^3 x dx = secx tanx + ∫secx dx
= secx tanx + ln(secx tanx)

∫sec^3 x dx = 1/2 (secx tanx + ln(secx + tanx)

plug in 0 and pi/4 to get

1/2 (√2*1 + ln(√2+1)) - 1/2(1*0 + ln(1+0))
= 1/2 (√2 + ln(√2+1))

multiply by 6 to get 3(√2 + ln(√2+1))

double check my math, and you will either
(a) see your mistake
(b) see my mistake