Posted by kajri on Tuesday, October 16, 2012 at 12:40am.
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
Calculus integral - Reiny, Tuesday, October 16, 2012 at 8:53am
According to the Wolfram integrator , this looks like a messy integration
looks like multiple integration by parts, arhhh!
Calculus integral - Steve, Tuesday, October 16, 2012 at 10:51am
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
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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