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March 6, 2015

March 6, 2015

Posted by **kajri** on Tuesday, October 16, 2012 at 12:40am.

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)

- Calculus integral -
**Reiny**, Tuesday, October 16, 2012 at 8:53amAccording 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!

- Calculus integral -
**Steve**, Tuesday, October 16, 2012 at 10:51amu = 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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