text: # 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 $-\frac{1}{3} |\ln \, \cos \frac{x}{3}| + C$
  
  • The answer in the book is wrong . You have the correct answer Jun 1, 2018 at 15:09
  • Another way to see the book is wrong is to substitute $y=x/3$, so the integral becomes $3\tan y dy$. So whatever that integrates to, the predictor must be $3$, not $1/3$.
  – J.G.
  Jun 1, 2018 at 15:32
  
  Your book is wrong! As a check, $$\frac{d}{dx}\left(-\frac13\bigg|\ln\cos\frac x3\bigg|\right)=-\frac1{3\cos\frac x3}\cdot\left(-\frac13\sin\frac x3\right)=\color{red}{\frac19}\tan\frac x3\neq \tan\frac x3.$$
  
  (...) and get the final answer $$-3 |\ln \, \cos \frac{x}{3}| + C$$
  
  But the answer in the back of the book is $$-\frac{1}{3} |\ln \, \cos \frac{x}{3}| + C$$
  
  You can differentiate to verify but you are right and the book is wrong!
  
  Just to make your life a bit simpler without this fraction $\frac{x}{3}$, just use u-sub.
  
  Put: $\frac{x}{3}=u$
  
  Then $\int tan(\frac{x}{3})dx= 3\int tan\ u\ du= 3$ln |sec u| +C$=$3 ln|sec($\frac{x}{3})|+C$
  
  At a glance, you're correct, because you'll divide by the $\frac 13$when you take it out of the trig function, not multiply by it.
  
  However, this formula sheet states that $$\int{\tan(x)dx}=\ln|\sec(x)|+C$$ (see the $C4$ section - page $9$)
  
  Your book is wrong as this is a standard integral and the value is:
  
  -3 ln|cos(x/3)| + C.
  
  If you want verification, differentiate this and u will get
  
  tan(x/3)
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