# Is this done correctly?

• Sep 2nd 2007, 05:26 PM
circuscircus
Is this done correctly?
$\displaystyle = \int [sec(\frac{x}{2}) + tan(\frac{3x+1}{5} )+ tan(7x)] dx$

$\displaystyle = \int sec(\frac{x}{2}) + \int tan(\frac{3x+1}{5} ) + \int tan(7x) dx$

$\displaystyle = 2ln|sec\frac{x}{2} + tan\frac{x}{2}| + \int tan(\frac{3x+1}{5} )+ \int tan(7x) dx$

$\displaystyle = 2ln|sec\frac{x}{2} + tan\frac{x}{2}| + (\frac{5}{3})(-ln|cos \frac {3}{5} | ) + (\frac{1}{7})(-ln|cos 7|)$

$\displaystyle = 2ln|sec\frac{x}{2} + tan\frac{x}{2}| -\frac{5}{3}ln|cos \frac {3}{5} | -\frac{1}{7}ln|cos 7|$

I don't have an answer to check but could someone give a thumbs up if they don't see anything wrong with it?
• Sep 2nd 2007, 05:38 PM
Jhevon
Quote:

Originally Posted by circuscircus
$\displaystyle = \int [sec(\frac{x}{2}) + tan(\frac{3x+1}{5} )+ tan(7x)] dx$

$\displaystyle = \int sec(\frac{x}{2}) + \int tan(\frac{3x+1}{5} ) + \int tan(7x) dx$

$\displaystyle = 2ln|sec\frac{x}{2} + tan\frac{x}{2}| + \int tan(\frac{3x+1}{5} )+ \int tan(7x) dx$

$\displaystyle = 2ln|sec\frac{x}{2} + tan\frac{x}{2}| + (\frac{5}{3})(-ln|cos \frac {3}{5} | ) + (\frac{1}{7})(-ln|cos 7|)$

$\displaystyle = 2ln|sec\frac{x}{2} + tan\frac{x}{2}| -\frac{5}{3}ln|cos \frac {3}{5} | -\frac{1}{7}ln|cos 7|$

I don't have an answer to check but could someone give a thumbs up if they don't see anything wrong with it?

it should be $\displaystyle 2 \ln \left| \sec \left( \frac {x}{2} \right) + \tan \left( \frac {x}{2} \right) \right| - \frac {5}{3} \ln \left| \cos \left( \frac {3x + 1}{5} \right) \right| - \frac {1}{7} \ln \left| \cos ( 7x ) \right| + C$