Question

\[\int\cot (x)\cdot\text dx\]

Answer 1

what is the easy way to do this?

Answer 2

put cot x = cos x/ sin x

Answer 3

yes please wait

Answer 4

\[\int\frac{\cos x}{\sin x}\text dx\]

Answer 5

\[\ln \left| \sin(x)\right|\]

Answer 6

+C !

Answer 7

\[\large{\int{\frac{cosx}{sinx}}dx}\]
\[\large{du = cos(x)dx}\]
right?

Answer 8

oh yeah,

Answer 9

du = cos(x) and sin (x) = u

Answer 10

\[\large{\int{\frac{cosx}{sinx}}=\int{\frac{du}{u}}}\]
Can u solve this integral now?

Answer 11

You will get : \(\large{ln|u| +C}\)

Answer 12

u = sin x

Answer 13

\[\int \cot x \text dx \]\[=\int\frac{\cos x}{\sin x}\text dx\] \(u=\sin (x)\) \(\text du = \cos(x)\text dx\) \[=\int\frac{\text du}u\]\[=\ln |u|+c\]\[=\ln |\sin x|+c\]

Answer 14

good work .. you got it @UnkleRhaukus

Answer 15

you are not familiar with this ?!
\[\int\limits_{}^{}\frac{u'(x)}{u(x)}dx = \ln(\left| u(x) \right|+C\]

Answer 16

Great!!! ..

Answer 17

thanks mathslover

Answer 18

@Neemo being familiar with that and being familiar with the proof are diff. things :)

Answer 19

yeah it makes sense now it just wasn't an obvious substitution

Answer 20

best of luck ..

have to go now

bbye

thanks!!

Answer 21

thank you