Calculus integral
posted by kajri .
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)

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! 
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
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