A little confused. Indefinite integral help. $\displaystyle\int\tan(5x+1)\,\mathrm{d}x$
Okay, so Im having trouble with this problem. I get the answer as $-\frac15\ln(\cos(5x+1))+C$ every time I do this question but It says it is incorrect.
 A: Hint:
$$\int \tan{x} \, dx = \int \frac{\sin{x}}{\cos{x}} dx  = -\int \frac{1}{\cos{x}} \frac{d \cos{x}}{dx}  \; dx$$
Also:
$$\int \frac{f'(x)}{f(x)} dx = \ln{f(x)} + C$$
In this case, you can substitute $u=5 x+1$, $x=(u-1)/5$, $dx=du/5$, so you integral becomes
$$\frac{1}{5} \int \tan{u} \, du = -\frac{1}{5} \ln{\cos{u}} + C = -\frac{1}{5} \ln{\cos{(5 x+1)}} + C$$
A: Some bot/program writers (what a yucky thing indeed!) are very picky, and since
$$\int\frac{1}{x}dx=\log|x|+C$$
(note the absolute value!), perhaps they want you to put the argument of the logarithm inside absolute value (and, of course, perhaps remarking it is non zero and etc.)... , i.e
$$\int\tan(5x+1)\,dx=-\frac{1}{5}\int\frac{-5\sin(5x+1)\,dx}{\cos(5x+1)}=:I$$
and since $\,-5\sin(5x+1)=d((\cos(5x+1))\,$ , we get at once (and on purpose I'll write in in several ways so that you try them)
$$I=-\frac{1}{5}\log|\cos(5x+1)|+C=\frac{1}{5}\log\left|\frac{1}{\cos(5x+1)}\right|+C=$$
$$=\frac{1}{5}\log|\sec(5x+1)|+C=\log\left|\frac{1}{\sqrt[5]{\cos(5x+1)}}\right|+C=$$
$$=\log\left|\cos^{-1/5}(5x+1)\right|+C=\log\left|\sqrt[5]{\sec(5x+1)}\right|+C=\ldots\,\,etc.$$
And then try all the above (and some others) with $\,\ln\,$ instead $\,\log\,$ (though the latter is more common, I think, in advanced mathematics), and at the end, no matter what the result is, you people should try to take all those who came up with that ridiculous idea of bot-programs in university level to court and accuse them of torture, mistreat and cruelty (go for a life-in-jail-without-parole sentence).
Anyway, and bottom line, trying to be honest and with good will: your answer is correct.
