Question

find indefinite integral of csc^2(t)/cot(t) dx
plz help!!!

Answer 1

Remember:
(1) d/dt [cot t] = -csc^2 t
(2) sin 2x = 2 sin x cos x

By (1) we see that INT {d/dt [cot t]} dt = INT {-csc^2 t} dt so cot t = - INT {csc^2 t} dt so INT {csc^2 t} dt = -cot t.

a) INT {6(csc t)^2 - 4e^t} dt
6(-cot t) - 4e^t + C

b) INT {5 sin 2x / sin x} dx
INT {5 * 2sin x cos x / sin x} dx
INT {10 cos x} dx
10 sin x + C

Answer 2

@angelarios did that clear everything for you?

Answer 3

im a little confused :/ @Gracy_KSL

Answer 4

what are you confused about?

Answer 5

I don't know where you're getting the very first part :o @Gracy_KSL

Answer 6

what do u mean why? or dt? @Loser66

Answer 7

oh yea sorry I meant dt :p sorry :( @Loser66

Answer 8

makes sense?

Answer 9

yes but I don't know where you are getting the cos/sin :o
isn't it when you multiply 1/sin^2 by cos/sin isn't it cos/sin^3 ??

Answer 10

your problem is \[\large\int \dfrac{csc^2 t}{cot t}dt\] right?

Answer 11

yes

Answer 12

numerator \(csc ^2t = \dfrac{1}{sin^2t}\) ok?

Answer 13

ok

Answer 14

denominator \(cot t= \dfrac{cos t}{sin t}\) right?

Answer 15

yes and after that you do numerator times reciprocal of denominator right??

Answer 16

yes, that's the way I do, but I had some mistake on the process, you can do the next step