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

u = 1 + cos x
du = -sin x dx

so we have
-du/u
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?
-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
Yes, you have it.
awesome, thanks man!
You are welcome.
Similar Questions
  1. That's the same as the integral of sin^2 x dx.Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx
    1. answers icon 0 answers
    1. answers icon 0 answers
  2. Evaluate the integral of (sin 2x)/(1+cos^2 x)1. u=cosx and du=-sinx *dx 2. evaluate the integral of -1/(1+u^2)*du 3. result is
    1. answers icon 1 answer
    1. answers icon 1 answer
more similar questions