u = 1 + cos x
du = -sin x dx
so we have
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
u = cosx
-1 du = sinx dx
so: integral -1 du/(1+u)^2
if x = 0, then u = 1
if x = pi/3, then u = 1
so 1/1+1 - 1/1+1 = 0
Is this correct?
-du/u^2 ---> -1/u + c
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?
-1/(1+cos x) + c
Yes, you have it.
awesome, thanks man!
You are welcome.
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