Evaluate indefinite integral $\int \tan(\frac{x}{3}) \, dx$ I need help verifying why I am getting an incorrect answer for the question evaluate the integral 
$$\int \tan\left(\frac{x}{3}\right) \, dx$$
I simplify the above equation using trig identities to get
$$\int \frac {\sin \left(\frac{x}{3}\right)}{\cos\left(\frac{x}{3}\right)} \, dx$$
I use the substitution method to find 
$$ du = -\frac{1}{3} \sin(x/3) \, dx$$ and so $dx = \frac{-3\,du}{\sin\frac{x}{3}}$
I plug the $u$ back into equation
$$ \int \frac {\sin\left(\frac{x}{3}\right)}{u} \cdot\frac {-3\,du}{\sin \left(\frac{x}{3}\right)}$$
I cross out the $\sin \left(\frac{x}{3}\right)$ and (this is where I may be going wrong), I pull out the $-3$ to be in front of the integral sign since it is a constant and solve for
$$-3 \int \frac{1}{u} \, du$$ and get the final answer $$ -3 \biggl|\,\ln \, \cos \frac{x}{3}\biggr| + C $$
But the answer in the back of the book is $ -\frac{1}{3} |\ln \, \cos \frac{x}{3}| + C $

Your book is wrong! As a check, $$\frac{d}{dx}\left(-\frac13\bigg|\ln\cos\frac x3\bigg|\right)=-\frac1{3\cos\frac x3}\cdot\left(-\frac13\sin\frac x3\right)=\color{red}{\frac19}\tan\frac x3\neq \tan\frac x3.$$
