what is the integration of $\int\cot(e^x)\cdot e^x dx$? what is
$$
\int\cot(e^x)\cdot e^x dx
$$
This is my answer is it right
$$
u=e^x\\
du=e^x dx\\
dx=\frac{1}{e^x}du
$$
Then $$\int\cot(u)\cdot u \frac{1}{u}du=\int \cot(u)\ du=\csc^2(u)$$
Is this answer right
 A: $$\int e^x\cot{e^x}dx=\int\frac{e^x\cos{e^x}}{\sin{e^x}}dx=\int\frac{1}{\sin{e^x}}d(\sin{e^x})=\ln|\sin{e^x}|+C$$
A: You have the right idea. Using $u=e^x$, we have $du=e^x\,dx$ so $$\int\cot(e^x)\cdot e^x\,dx=\int\cot u \, du=\int\frac{\cos u}{\sin u}\, du$$ Now let $v=\sin u$. Then $dv=\cos u\, du$ so $$\int\cot u\, du=\int\frac1v\, dv=\ln|v|+C$$ where $C$ is a constant. 
Therefore $$\boxed{\int\cot(e^x)\cdot e^x\,dx=\ln|\sin(e^x)|+C}$$
A: No, your answer is not correct, but you were on the right track for the most part. Your problem was that you were looking at the wrong table when you were taking the integral of $\cot{u}$. Instead of the antiderivate of the cotangent function which is $\int\cot{x}\:dx=\ln{|\sin{x}|}+C$, you used its derivative $\left(\frac{d}{dx}\cot{x}=-\csc^2 x\right)$ to finish off the problem (you also dropped the minus sign for some reason). You got it completely backwards. Anyway, this is a simple integration by substitution (sometimes called u-substitution) problem. Integration by substitution is nothing more than the chain rule done backwards.
$$
\int\cot e^x\cdot e^x dx=\\
u=e^{x}\implies\\
\frac{du}{dx}=\frac{d}{dx}e^{x}\implies\\
\frac{du}{dx}=e^{x}\implies\\
dx=\frac{du}{e^{x}}\\
=\int\cot u\cdot u \frac{du}{u}=
\int\cot{u} du=
\ln{|\sin{u}|}+C=
\ln{|\sin{e^{x}}|}+C
$$
Answer: $\ln{|\sin{e^{x}}|}+C$.
