How do you integrate $int thetasecthetatantheta$ by parts?

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Answer 1 · 2017-02-04T08:44:28

$thetasectheta-ln|sectheta+tantheta|+C$

the parts formula

$intuv'd theta=uv-intvu'd theta$

$u=theta=>u'=1$

$v'=secthetatantheta=>v=sectheta$

$I=intuv'd theta=uv-intvu'd theta=thetasectheta-intsecthetad theta$

the problem now is to integrate $intsecthetad theta$

proceed as follows

$intsecthetad theta=int((sectheta)xx((sectheta+tantheta)/(sectheta+tantheta)))d theta$

$int((sec^2theta+secthetatantheta)/(sectheta+tantheta))d theta$

the numerator is the derivative of the denominator, so we have a log integral

$:.intsecthetad theta=ln|sectheta+tantheta|$

the original integral is now complete

$I=thetasectheta-intsecthetad theta$

$=>I=thetasectheta-ln|sectheta+tantheta|+C$

How do you integrate int thetasecthetatanthetaint thetasecthetatantheta by parts?