u = 1 + cos x
du = -sin x dx
so we have
-du/u
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
9 answers
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?
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?
-du/(u)^2
-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?
so
-1/(1+cos x) + c
-1/(1+cos x) + c
Yes, you have it.
awesome, thanks man!
You are welcome.