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August 30, 2014

August 30, 2014

Posted by **Paul** on Tuesday, August 7, 2012 at 5:09pm.

integral 1 = a and b = 4 of

sinx dx/(1+cos)^2

u = cosx

du = -sinx dx

so from here I don't know if I can do:

-1 du = sinx dx

or 1/sin du = x dx

- Calculus -
**Damon**, Tuesday, August 7, 2012 at 5:24pmu = 1 + cos x

du = -sin x dx

so we have

-du/u

- Calculus -
**Paul**, Tuesday, August 7, 2012 at 5:25pmso I tried again and I got this, don't know if it's good:(I also put the wrong a and b)

integral 0 = a and b = pi/3 of

sinx dx/(1+cos)^2

u = cosx

du= -sinxdx

-1 du = sinx dx

so: integral -1 du/(1+u)^2

= 1/(u+1)

if x = 0, then u = 1

if x = pi/3, then u = 1

so 1/1+1 - 1/1+1 = 0

Is this correct?

- I mean -
**Damon**, Tuesday, August 7, 2012 at 5:26pm-du/(u)^2

- Calculus -
**Damon**, Tuesday, August 7, 2012 at 5:27pm-du/u^2 ---> -1/u + c

- Calculus -
**Paul**, Tuesday, August 7, 2012 at 5:30pmok Damon, so if you integrate what you have me, you get 1/u no? Because 1/u^2 = -1/u, so -1/u^2 = 1/u?

- Calculus -
**Damon**, Tuesday, August 7, 2012 at 5:30pmso

-1/(1+cos x) + c

- Calculus -
**Damon**, Tuesday, August 7, 2012 at 5:30pmYes, you have it.

- Calculus -
**Paul**, Tuesday, August 7, 2012 at 5:31pmawesome, thanks man!

- Calculus -
**Damon**, Tuesday, August 7, 2012 at 5:31pmYou are welcome.

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