Question

Question......

Answer 1

\[\int\limits_{}^{} \frac{Cosx }{ Sin ^{2}x }\]

Answer 2

I got -1/sinx +C

Answer 3

@misty1212

Answer 4

@kohai

Answer 5

@freckles

Answer 6

find the antiderivative

Answer 7

a sub for sin(x) will do the trick
let me check your solution.,..

\[\int\limits_{}^{}\frac{du}{u^2}= \int\limits u^{-2} du=\frac{u^{-1}}{-1}+C\]
looks great

Answer 8

you can also check your answer by differentiating your answer

Answer 9

\[[-(\sin(x))^{-1}]' \\=-(-1)(\sin(x))^{-1-1} \cdot \cos(x) \\ =(\sin(x))^{-2} \cos(x) =\frac{\cos(x)}{\sin^2(x)}\]

Answer 10

ok

Answer 11

thank you very much