Question

use the derivative of cos^2(x) to find the integral:

Answer 1

and where's the integral ?

Answer 2

\[\int\limits_{}^{}\frac{ \sin2x }{ \cos^2(x) + 9 }\]

Answer 3

WHOOPS forgot the dx

Answer 4

aaaaaand, i think i've got the ans. But if anyone else has it, i'd like to check anyways :P

Answer 5

sin2x = 2sinx cosx
cos x =t
-sin x dx =dt
then substitute t^2 as v
You get 2t as dv

Answer 6

nono we don't do that, we put a minus sign inside and outside the integral sign, and then we substitute -sin2x dx for dy, if we say that cos^2(x) = y

Answer 7

and then the denominator is y + 9 = y +3^2, so it becomes cot'(y) + c

Answer 8

it's the same thing however

Answer 9

you'll get the answer in log
not in cot x

Answer 10

no we wont, the derivative of cotINVERSEx is -1/1+x^2 so it'll be that into 1/3

Answer 11

you'll get -2t/t^2 + 9 as your integral after substituting cos x = t
and t/t^2 + 9 is not cot inverse x

Answer 12

it's not t, it's dt.

Answer 13

ughh how do i explain this poor fellow
well dude, why do you ask questions here when you know you'll get the correct answer ?
i've checked it twice and thrice and what i'm saying is the answer

Answer 14

ok, ok, ok, i got it, sorry. My bad.

Answer 15

fine