Posted by Paul on Tuesday, August 7, 2012 at 5:09pm.
Evaluate the following integral
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 5:24pm
u = 1 + cos x
du = sin x dx
so we have
du/u

Calculus  Paul, Tuesday, August 7, 2012 at 5:25pm
so 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 5:30pm
ok 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 5:30pm
so
1/(1+cos x) + c

Calculus  Damon, Tuesday, August 7, 2012 at 5:30pm
Yes, you have it.

Calculus  Paul, Tuesday, August 7, 2012 at 5:31pm
awesome, thanks man!

Calculus  Damon, Tuesday, August 7, 2012 at 5:31pm
You are welcome.
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