calculus help: integral of cot^2 (x)?

Question

Please help. i tried u substitution. but i get confused. i'd do u=tanx then du=sec^2(x) and it doesnt work out. can i get some help?

Answer 1

Note, cot^2 (x) = cosec^2 x - 1. Also, d/dx (cot x) = -cosec^2 x.

So ∫ (cot^2 x dx)
= ∫ (cosec^2 x dx) - ∫ 1 dx
= -cot x - x + c.

If you subsitute u = tan x, du = sec^2 x dx = (u^2 + 1) dx you should get
∫ (1/u^2 . 1/(u^2+1) du)
You need to use partial fractions to separate this. In this case it turns out to have a nice form:
= ∫ (1/u^2 - 1/(u^2+1)) du
(over the denominator u^2(u^2+1) this gives (u^2+1)-u^2 = 1, which is what we need.)
= ∫ u^(-2) du - ∫ 1/(u^2+1) du
= -u^(-1) - arctan u + c
= -1/tan x - arctan (tan x) + c
= -cot x - x + c.

Answer 2

Remember that

cot^2(x) = csc^2(x) - 1

Therefore

Integral (cot^2(x)) dx =

Integral (csc^2(x) - 1)dx

Which we can now solve directly, since -csc^2(x) is a known derivative (in fact, it's the derivative of cot(x).

-cot(x) - x + C

Answer 3

well, of course i get cot(x)^2 when i differentiate......
the question is how do i solve the integral without knowing that it's equal to -x - cot(x)

if i use cos(x)^2 + sin(x)^2 == 1 i get to solve
Integral of 1/sin(x)^2 dx
which i don't find easier

http://www.sarcasticgazette.com

Answer 4

Integrate ∫cot²(x)dx.

∫cot²(x)dx = ∫{csc²(x) - 1}dx = -cot(x) - x + C