Question

I am doing an integral of cot^2(x).
If I get stuck pls save me. I have different ideas.... guide, but don't do my problem. Please...!

Answer 1

\[\Large \int\limits_{ }^{ }\cot^2(x)~dx\]\[\Large \int\limits_{ }^{ }\csc^2(x)\cos^2(x)~dx\]\[\Large \int\limits_{ }^{ }\left(\begin{matrix}\cot^2(x)+1 \\ \end{matrix}\right) \cos^2(x)~dx\]\[\Large \int\limits_{ }^{ } \cot^2(x) \cos^2(x)+\cot^2(x)~dx \]
So so far I get,
\[\Large \int\limits_{ }^{ }\cot^2(x)~dx =\Large \int\limits_{ }^{ } \cot^2(x) \cos^2(x)+\cot^2(x)~dx \]\[\Large \int\limits_{ }^{ }\cot^2(x)~dx =\Large \int\limits_{ }^{ } \cot^2(x) \cos^2(x) dx+\Large \int\limits_{ }^{ } \cot^2(x) ~dx\]

Answer 2

Does this get me any where, or I should try something else? Kainui can you tell me?

Answer 3

I knew I am going to sink, @Kainui can you give me a hint?

Answer 4

Ohh... I know

Answer 5

Yeah, I would suggest you use this identity: \[\LARGE \frac{\sin^2 \theta}{\sin^2 \theta}+\frac{\cos^2 \theta}{\sin^2 \theta}=\frac{1}{\sin^2 \theta}\] So all I've done is divided the pythagorean identity by sin^2theta.

Answer 6

\[\Large \int\limits_{ }^{ } \cot^2(x)dx=0\]\[\Large \int\limits_{ }^{ }( \csc^2(x)-1 )dx=0\]\[\Large \int\limits_{ }^{ } \csc^2(x) dx-\Large \int\limits_{ }^{ } 1~dx\]

Answer 7

\[\Large -\cot(x)+x+C\]

Answer 8

I used the rule ina wrong place... but tnx for suggesting, Kainiu. I for the least part knew that this is the rule:)

Answer 9

LOL how would I be so off....

Answer 10

Yeah, you got it haha.

Answer 11

tnx again!

Answer 12

You pretty much did all the work, I think you figured it out just in time. =P