What is wrong in my answer? Subject: finding the integral of $\cot x$

Question

$$\int \cot x \,dx=\int \frac{\cos x\sin x}{\sin^2 x}\,dx. $$ Assume $t= \sin^2 x$. Then $dt= 2\sin x\cos x \,dx$. Using the identity: $$ \sin 2x =2\sin x\cos x$$

$$\int \cot x \,dx=\int \frac{1}{2t}\,dt=\frac{1}{2}\ln(\sin x)^2+c $$ Can you help me find out what I did wrong? Thanks a lot!

Answer 1

Note that

$$\frac12\ln(\sin(x))^2+c=\ln|\sin(x)|+c$$

Answer 2

You can do even less: notice that $\cot x = \frac{\cos x}{\sin x}$ and that the derivative of $\sin$ is $\cos$. Hence $$\int \cot x \,dx = \int \frac{\cos x}{\sin x}\,dx = \log \lvert \sin x \rvert + c, c \in \mathbb{R}.$$

Answer 3

Your integrand is correct:

$$\int \cot x\, dx = \int \frac{1}{2} \frac{dt}{t} = \frac{1}{2} \log\lvert t \rvert + C.$$ The issue is just with your final substitution: $$\frac{1}{2} \log\lvert t\rvert + C = \frac{1}{2} \log( \sin^2x) + C = \log\lvert \sin(x)\rvert + C.$$

Answer 4

your work is correct. $$\int \cot x \,dx=\int \frac{1}{2t}\,dt=\frac{1}{2}\ln(\sin x)^2+c =\ln\lvert \sin x\rvert +c.$$

I am surprised that you did not make the integration simpler by,

$$\int \cot x \,dx=\int \frac{\cos x}{\sin x}\,dx = \int \frac{du}{u} =\ln\lvert \sin x\rvert +c. $$